MATH 204 Introduction to Statistics

MATH 204 Introduction to Statistics
Lecture 12: Further Inference
JMG
1

Goals for LectureIn this lecture, we introduce statistical inference for
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Goals for Lecture

In this lecture, we introduce statistical inference for
- a single proportion (6.1)

Goals for Lecture

In this lecture, we introduce statistical inference for
- a single proportion (6.1)
- a difference of two proportions (6.2)

Goals for Lecture

In this lecture, we introduce statistical inference for
- a single proportion (6.1)
- a difference of two proportions (6.2)
- one-sample means (7.1)

Goals for Lecture

In this lecture, we introduce statistical inference for
- a single proportion (6.1)
- a difference of two proportions (6.2)
- one-sample means (7.1)
- paired data (7.3)

Goals for Lecture

In this lecture, we introduce statistical inference for
- a single proportion (6.1)
- a difference of two proportions (6.2)
- one-sample means (7.1)
- paired data (7.3)
- a difference of two means (7.3)

Goals for Lecture

In this lecture, we introduce statistical inference for
- a single proportion (6.1)
- a difference of two proportions (6.2)
- one-sample means (7.1)
- paired data (7.3)
- a difference of two means (7.3)
After we discuss these topics along with power calculations and simple ordinary least squares regression (7.4 and Chapter 8), we will return to discuss goodness of fit (6.3) and testing for independence (6.4).

Learning ObjectivesAfter this lecture, you should be able to conduct and apply point estimates, interval estimates, and hypothesis tests for
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Learning Objectives

After this lecture, you should be able to conduct and apply point estimates, interval estimates, and hypothesis tests for
- proportions, difference of proportions, one sample means, paired data, and difference of two means.

Learning Objectives

After this lecture, you should be able to conduct and apply point estimates, interval estimates, and hypothesis tests for
- proportions, difference of proportions, one sample means, paired data, and difference of two means.
This includes knowing when to use which type of hypothesis test, and

Learning Objectives

After this lecture, you should be able to conduct and apply point estimates, interval estimates, and hypothesis tests for
- proportions, difference of proportions, one sample means, paired data, and difference of two means.
This includes knowing when to use which type of hypothesis test, and
what is the appropriate R command(s) to use.

Learning Objectives

After this lecture, you should be able to conduct and apply point estimates, interval estimates, and hypothesis tests for
- proportions, difference of proportions, one sample means, paired data, and difference of two means.
This includes knowing when to use which type of hypothesis test, and
what is the appropriate R command(s) to use.
Many of our interval estimates and hypothesis tests will rely on a central limit theorem. We quickly review these.

Central Limit Theorems

Recall our CLT for sample proportions:

When observations are independent and sample size is sufficiently large, then the sampling distribution for the sample proportion $^p$ is approximately normal with mean $μ_{^p} = p$ and standard error $S E_{^p} = \sqrt{\frac{p (1 - p)}{n}}$ .

Central Limit Theorems

Recall our CLT for sample proportions:

When observations are independent and sample size is sufficiently large, then the sampling distribution for the sample proportion $^p$ is approximately normal with mean $μ_{^p} = p$ and standard error $S E_{^p} = \sqrt{\frac{p (1 - p)}{n}}$ .

We also have a CLT for the sample mean:

When we collect samples of sufficiently large size $n$ from a population with mean $μ$ and standard deviation $σ$ , then the sampling distribution for the sample mean $¯ x$ is approximately normal with mean $μ_{¯ x} = μ$ and standard error $S E_{¯ x} = \frac{σ}{\sqrt{n}}$ .

Central Limit Theorems

Recall our CLT for sample proportions:

When observations are independent and sample size is sufficiently large, then the sampling distribution for the sample proportion $^p$ is approximately normal with mean $μ_{^p} = p$ and standard error $S E_{^p} = \sqrt{\frac{p (1 - p)}{n}}$ .

We also have a CLT for the sample mean:

When we collect samples of sufficiently large size $n$ from a population with mean $μ$ and standard deviation $σ$ , then the sampling distribution for the sample mean $¯ x$ is approximately normal with mean $μ_{¯ x} = μ$ and standard error $S E_{¯ x} = \frac{σ}{\sqrt{n}}$ .

Later we will return to discuss precisely what we mean by "samples of sufficiently large size."

Central Limit Theorems

Recall our CLT for sample proportions:

When observations are independent and sample size is sufficiently large, then the sampling distribution for the sample proportion $^p$ is approximately normal with mean $μ_{^p} = p$ and standard error $S E_{^p} = \sqrt{\frac{p (1 - p)}{n}}$ .

We also have a CLT for the sample mean:

When we collect samples of sufficiently large size $n$ from a population with mean $μ$ and standard deviation $σ$ , then the sampling distribution for the sample mean $¯ x$ is approximately normal with mean $μ_{¯ x} = μ$ and standard error $S E_{¯ x} = \frac{σ}{\sqrt{n}}$ .

Later we will return to discuss precisely what we mean by "samples of sufficiently large size."

We use sampling distributions for obtaining confidence intervals and conducting hypothesis tests.

Videos for CLTCentral limit theorems are reviewed in the videos included on the next few slides.
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CLT Video 1

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CLT Video 2

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CLT Video 3

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CI for a ProportionOnce you've determined a one-proportion CI would be helpful for an application, there are four steps to constructing the interval:
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CI for a Proportion

Once you've determined a one-proportion CI would be helpful for an application, there are four steps to constructing the interval:
- Identify the point estimate $^p$ and the sample size $n$ , and determine what confidence level you want.

CI for a Proportion

Once you've determined a one-proportion CI would be helpful for an application, there are four steps to constructing the interval:
- Identify the point estimate $^p$ and the sample size $n$ , and determine what confidence level you want.
- Verify the conditions to ensure $^p$ is nearly normal, that is, check that $n^p \geq 10$ and $n (1 -^p) \geq 10$ .

CI for a Proportion

Once you've determined a one-proportion CI would be helpful for an application, there are four steps to constructing the interval:
- Identify the point estimate $^p$ and the sample size $n$ , and determine what confidence level you want.
- Verify the conditions to ensure $^p$ is nearly normal, that is, check that $n^p \geq 10$ and $n (1 -^p) \geq 10$ .
- If the conditions hold, estimate the standard error $S E$ using $^p$ , find the appropriate $z^{*}$ , and compute $^p \pm z^{*} \times S E$ .

CI for a Proportion

Once you've determined a one-proportion CI would be helpful for an application, there are four steps to constructing the interval:
- Identify the point estimate $^p$ and the sample size $n$ , and determine what confidence level you want.
- Verify the conditions to ensure $^p$ is nearly normal, that is, check that $n^p \geq 10$ and $n (1 -^p) \geq 10$ .
- If the conditions hold, estimate the standard error $S E$ using $^p$ , find the appropriate $z^{*}$ , and compute $^p \pm z^{*} \times S E$ .
- Interpret the CI in the context of the problem.

Proportion CI ExampleConsider the following problem:
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Proportion CI Example

Consider the following problem:

Is parking a problem on campus? A randomly selected group of 89 faculty and staff are asked whether they are satisfied with campus parking or not. Of the 89 individuals surveyed, 23 indicated that they are satisfied. What is the proportion $p$ of faculty and staff that are satisfied with campus parking?

Proportion CI Example

Consider the following problem:

Is parking a problem on campus? A randomly selected group of 89 faculty and staff are asked whether they are satisfied with campus parking or not. Of the 89 individuals surveyed, 23 indicated that they are satisfied. What is the proportion $p$ of faculty and staff that are satisfied with campus parking?

Explain why this problem is suitable for a one-proportion CI. Is this CLT applicable? If so, obtain a 95% CI for a point estimate.

Proportion CI VideoConfidence intervals for a proportion are reviewed in this video:

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Hypothesis Test for a ProportionOnce you've determined a one-proportion hypothesis test is the correct test for a problem, there are four steps to completing the test:
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Hypothesis Test for a Proportion

Once you've determined a one-proportion hypothesis test is the correct test for a problem, there are four steps to completing the test:
- Identify the parameter of interest, list hypotheses, identify the significance level, and identify $^p$ and $n$ .

Hypothesis Test for a Proportion

Once you've determined a one-proportion hypothesis test is the correct test for a problem, there are four steps to completing the test:
- Identify the parameter of interest, list hypotheses, identify the significance level, and identify $^p$ and $n$ .
- Verify conditions to ensure $^p$ is nearly normal under $H_{0}$ . For one-proportion hypothesis tests, use the null value $p_{0}$ to check the conditions $n p_{0} \geq 10$ and $n (1 - p_{0}) \geq 10$ .

Hypothesis Test for a Proportion

Once you've determined a one-proportion hypothesis test is the correct test for a problem, there are four steps to completing the test:
- Identify the parameter of interest, list hypotheses, identify the significance level, and identify $^p$ and $n$ .
- Verify conditions to ensure $^p$ is nearly normal under $H_{0}$ . For one-proportion hypothesis tests, use the null value $p_{0}$ to check the conditions $n p_{0} \geq 10$ and $n (1 - p_{0}) \geq 10$ .
- If the conditions hold, compute the standard error, again using $p_{0}$ , compute the $Z$ -score, and identify the p-value.

Hypothesis Test for a Proportion

Once you've determined a one-proportion hypothesis test is the correct test for a problem, there are four steps to completing the test:
- Identify the parameter of interest, list hypotheses, identify the significance level, and identify $^p$ and $n$ .
- Verify conditions to ensure $^p$ is nearly normal under $H_{0}$ . For one-proportion hypothesis tests, use the null value $p_{0}$ to check the conditions $n p_{0} \geq 10$ and $n (1 - p_{0}) \geq 10$ .
- If the conditions hold, compute the standard error, again using $p_{0}$ , compute the $Z$ -score, and identify the p-value.
- Evaluate the hypothesis test by comparing the p-value to the significance level $α$ , and provide a conclusion in the context of the problem.

Hypotheses for One Proportion TestsThe null hypothesis for a one-proportion test is typically stated as H0:p=p0H0:p=p0 where p0p0 is a hypothetical value for pp. 
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Hypotheses for One Proportion Tests

The null hypothesis for a one-proportion test is typically stated as $H_{0} : p = p_{0}$ where $p_{0}$ is a hypothetical value for $p$ .
The corresponding alternative hypothesis is then one of

Hypotheses for One Proportion Tests

The null hypothesis for a one-proportion test is typically stated as $H_{0} : p = p_{0}$ where $p_{0}$ is a hypothetical value for $p$ .
The corresponding alternative hypothesis is then one of
- $H_{A} : p \neq p_{0}$ (two-sided),

Hypotheses for One Proportion Tests

The null hypothesis for a one-proportion test is typically stated as $H_{0} : p = p_{0}$ where $p_{0}$ is a hypothetical value for $p$ .
The corresponding alternative hypothesis is then one of
- $H_{A} : p \neq p_{0}$ (two-sided),
- $H_{A} : p < p_{0}$ (one-sided less than), or

Hypotheses for One Proportion Tests

The null hypothesis for a one-proportion test is typically stated as $H_{0} : p = p_{0}$ where $p_{0}$ is a hypothetical value for $p$ .
The corresponding alternative hypothesis is then one of
- $H_{A} : p \neq p_{0}$ (two-sided),
- $H_{A} : p < p_{0}$ (one-sided less than), or
- $H_{A} : p > p_{0}$ (one-sided greater than)

Proportion Hypothesis Test Example

Is parking a problem on campus? A randomly selected group of 89 faculty and staff are asked whether they are satisfied with campus parking or not. Of the 89 individuals surveyed, 23 indicated that they are satisfied.

Proportion Hypothesis Test Example

Is parking a problem on campus? A randomly selected group of 89 faculty and staff are asked whether they are satisfied with campus parking or not. Of the 89 individuals surveyed, 23 indicated that they are satisfied.

Let's study this problem using a hypothesis testing framework.

Proportion Hypothesis Test Example

Is parking a problem on campus? A randomly selected group of 89 faculty and staff are asked whether they are satisfied with campus parking or not. Of the 89 individuals surveyed, 23 indicated that they are satisfied.

Let's study this problem using a hypothesis testing framework.

Here is relevant R code:

## [1] 44.5 44.5

## [1] -4.557991

## [1] 5.164528e-06

p_hat <- 23/89
n <- 89
p0 <- 0.5
(c(n*p0,n*(1-p0))) # success-failure condition
SE <- sqrt((p0*(1-p0))/n)
(z_val <- (p_hat - p0)/SE)
(p_value <- 2*(pnorm(z_val)))

R Command(s) for Proportion Hypothesis TestIn cases where the CLT applies, we can use a normal distribution to directly compute p-values with the pnorm function. 
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R Command(s) for Proportion Hypothesis Test

In cases where the CLT applies, we can use a normal distribution to directly compute p-values with the pnorm function.
Alternatively, R has a built-in function prop.test that automates one-proportion hypothesis testing.

R Command(s) for Proportion Hypothesis Test

In cases where the CLT applies, we can use a normal distribution to directly compute p-values with the pnorm function.
Alternatively, R has a built-in function prop.test that automates one-proportion hypothesis testing.
So, in our parking problem example we would use

## 
##     1-sample proportions test without continuity correction
## 
## data:  23 out of 89, null probability 0.5
## X-squared = 20.775, df = 1, p-value = 5.165e-06
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
##  0.1788154 0.3580294
## sample estimates:
##        p 
## 0.258427

prop.test(23,89,p=0.5,correct = FALSE)

Problems to PracticeLet's take a minute to do some practice problems. 
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Problems to Practice

Let's take a minute to do some practice problems.
Try problems 6.7 and 6.13 from the textbook.

Problems Involving Difference of Proportions

Consider the following problem:

An online poll on January 10, 2017 reported that 97 out of 120 people in Virginia between the ages of 18 and 29 believe that marijuana should be legal, while 84 out of 111 who are 30 and over held this belief. Is there a difference between the proportion of young people who favor marijuana leagalization as compared to people who are older?

Problems Involving Difference of Proportions

Consider the following problem:

An online poll on January 10, 2017 reported that 97 out of 120 people in Virginia between the ages of 18 and 29 believe that marijuana should be legal, while 84 out of 111 who are 30 and over held this belief. Is there a difference between the proportion of young people who favor marijuana leagalization as compared to people who are older?

Think about how this problem is different than problems for a single proportion. The basic point here is that there are two groups, and we want to compare proportions across the two groups.

Problems Involving Difference of Proportions

Consider the following problem:

An online poll on January 10, 2017 reported that 97 out of 120 people in Virginia between the ages of 18 and 29 believe that marijuana should be legal, while 84 out of 111 who are 30 and over held this belief. Is there a difference between the proportion of young people who favor marijuana leagalization as compared to people who are older?

Think about how this problem is different than problems for a single proportion. The basic point here is that there are two groups, and we want to compare proportions across the two groups.

Data for the problem stated above might look as follows (only the first few rows are shown).

## # A tibble: 6 × 2
##   support_legalize age_group
##   <chr>            <fct>    
## 1 Yes              >30      
## 2 No               18-29    
## 3 Yes              >30      
## 4 Yes              >30      
## 5 Yes              >30      
## 6 Yes              18-29

Plotting Grouped Proportion Data

We can used grouped bar plots or a mosaic plot to visualize data related to grouped proportions:

Sampling Distribution for Difference of Proportions

The difference ${^p}_{1} - {^p}_{2}$ can be modeled using a normal distribution when

The data are independent within and between the two groups. Generally this is satisfied if the data come from two independent random samples or if the data come from a randomized experiment.
The success-failure condition holds for both groups, where we check successes and failures in each group separately.

Sampling Distribution for Difference of Proportions

The difference ${^p}_{1} - {^p}_{2}$ can be modeled using a normal distribution when

The data are independent within and between the two groups. Generally this is satisfied if the data come from two independent random samples or if the data come from a randomized experiment.
The success-failure condition holds for both groups, where we check successes and failures in each group separately.

When these conditions are satisfied, the standard error of ${^p}_{1} - {^p}_{2}$ is

$S E = \sqrt{\frac{p_{1} (1 - p_{1})}{n_{1}} + \frac{p_{2} (1 - p_{2})}{n_{2}}},$ where $p_{1}$ and $p_{2}$ represent the population proportions, and $n_{1}$ and $n_{2}$ represent the sample sizes.

CI for Difference of Proportions

To construct a CI for a difference of proportion, we can apply the formulas

$S E = \sqrt{\frac{p_{1} (1 - p_{1})}{n_{1}} + \frac{p_{2} (1 - p_{2})}{n_{2}}}, C I = point estimate \pm z^{*} \times S E$

CI for Difference of Proportions

To construct a CI for a difference of proportion, we can apply the formulas

$S E = \sqrt{\frac{p_{1} (1 - p_{1})}{n_{1}} + \frac{p_{2} (1 - p_{2})}{n_{2}}}, C I = point estimate \pm z^{*} \times S E$

If necessary, we can use the plug-in principle to estimate the standard error.

$S E \approx \sqrt{\frac{{^p}_{1} (1 - {^p}_{1})}{n_{1}} + \frac{{^p}_{2} (1 - {^p}_{2})}{n_{2}}}$

CI for Difference of Proportions ExamplesLet's obtain a 90% CI for the difference of proportions in the marijuana legalization question. Here is the relevant R code:
Result
Code
## [1] 97 23 84 27
## [1] -0.03775321  0.14090636
n1 <- 120
n2 <- 111
p1 <- 97/n1
p2 <- 84/n2
p_diff <- p1 - p2
(c(n1*p1,n1*(1-p1),n2*p2,n2*(1-p2))) # success-failure conditions
SE <- sqrt(((p1*(1-p1))/n1) + ((p2*(1-p2))/n2))
z_90 <- -qnorm(0.05) # find the appropriate z-value for 90% CI
(CI <- p_diff + z_90 * c(-1,1) * SE) # 90% CI
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CI for Difference of Proportions ExamplesLet's obtain a 90% CI for the difference of proportions in the marijuana legalization question. Here is the relevant R code:
Result
Code
## [1] 97 23 84 27
## [1] -0.03775321  0.14090636
n1 <- 120
n2 <- 111
p1 <- 97/n1
p2 <- 84/n2
p_diff <- p1 - p2
(c(n1*p1,n1*(1-p1),n2*p2,n2*(1-p2))) # success-failure conditions
SE <- sqrt(((p1*(1-p1))/n1) + ((p2*(1-p2))/n2))
z_90 <- -qnorm(0.05) # find the appropriate z-value for 90% CI
(CI <- p_diff + z_90 * c(-1,1) * SE) # 90% CI
Let's try some more examples together. 
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Hypotheses for Difference of Proportion TestsThe null hypothesis for a difference of proportions test is typically stated as H0:p1−p2=p0H0:p1−p2=p0 where p0p0 is a hypothetical value for pp. 
59

Hypotheses for Difference of Proportion Tests

The null hypothesis for a difference of proportions test is typically stated as $H_{0} : p_{1} - p_{2} = p_{0}$ where $p_{0}$ is a hypothetical value for $p$ .
The corresponding alternative hypothesis is then one of

Hypotheses for Difference of Proportion Tests

The null hypothesis for a difference of proportions test is typically stated as $H_{0} : p_{1} - p_{2} = p_{0}$ where $p_{0}$ is a hypothetical value for $p$ .
The corresponding alternative hypothesis is then one of
- $H_{A} : p_{1} - p_{2} \neq p_{0}$ (two-sided),

Hypotheses for Difference of Proportion Tests

The null hypothesis for a difference of proportions test is typically stated as $H_{0} : p_{1} - p_{2} = p_{0}$ where $p_{0}$ is a hypothetical value for $p$ .
The corresponding alternative hypothesis is then one of
- $H_{A} : p_{1} - p_{2} \neq p_{0}$ (two-sided),
- $H_{A} : p_{1} - p_{2} < p_{0}$ (one-sided less than), or

Hypotheses for Difference of Proportion Tests

The null hypothesis for a difference of proportions test is typically stated as $H_{0} : p_{1} - p_{2} = p_{0}$ where $p_{0}$ is a hypothetical value for $p$ .
The corresponding alternative hypothesis is then one of
- $H_{A} : p_{1} - p_{2} \neq p_{0}$ (two-sided),
- $H_{A} : p_{1} - p_{2} < p_{0}$ (one-sided less than), or
- $H_{A} : p_{1} - p_{2} > p_{0}$ (one-sided greater than)

Hypotheses for Difference of Proportion Tests

The null hypothesis for a difference of proportions test is typically stated as $H_{0} : p_{1} - p_{2} = p_{0}$ where $p_{0}$ is a hypothetical value for $p$ .
The corresponding alternative hypothesis is then one of
- $H_{A} : p_{1} - p_{2} \neq p_{0}$ (two-sided),
- $H_{A} : p_{1} - p_{2} < p_{0}$ (one-sided less than), or
- $H_{A} : p_{1} - p_{2} > p_{0}$ (one-sided greater than)
When $p_{0} = 0$ we need to use the pooled proportion.

Pooled Proportion

When the null hypothesis is that the proportions are equal, use the pooled proportion ${^p}_{pooled}$ , where

${^p}_{pooled} = \frac{number of "successes"}{number of cases} = \frac{{^p}_{1} n_{1} + {^p}_{2} n_{2}}{n_{1} + n_{2}}$

Pooled Proportion

When the null hypothesis is that the proportions are equal, use the pooled proportion ${^p}_{pooled}$ , where

${^p}_{pooled} = \frac{number of "successes"}{number of cases} = \frac{{^p}_{1} n_{1} + {^p}_{2} n_{2}}{n_{1} + n_{2}}$

The pooled proportion ${^p}_{pooled}$ is used to check the success-failure condition and to estimate the standard error.

Hypothesis Test for Difference of Proportions

Let's apply the hypothesis testing framework to this problem:

An online poll on January 10, 2017 reported that 97 out of 120 people in Virginia between the ages of 18 and 29 believe that marijuana should be legal, while 84 out of 111 who are 30 and over held this belief. Is there a difference between the proportion of young people who favor marijuana leagalization as compared to people who are older?

## [1] 69.73593 19.26407

## [1] 0.9510127

## [1] 0.3415979

n1 <- 120; n2 <- 111
p1 <- 97/120; p2 <- 84/111; p_diff <- p1 - p2
p_pooled <- (p1*n1+p2*n2)/(n1+n2)
(c(n*p_pooled,n*(1-p_pooled)))
SE <- sqrt((p_pooled*(1-p_pooled))/n1 + (p_pooled*(1-p_pooled))/n2)
(z_val <- (p_diff - 0.0)/SE)
(p_value <- 2*(1-pnorm(z_val)))

R Commands for Difference of Proportion TestThe R command prop.test also works for testing a difference of proportions. 
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R Commands for Difference of Proportion Test

The R command prop.test also works for testing a difference of proportions.
For example, we could solve the marijuana legalization problem with the following:

## 
##     2-sample test for equality of proportions without continuity correction
## 
## data:  c(97, 84) out of c(120, 111)
## X-squared = 0.90443, df = 1, p-value = 0.3416
## alternative hypothesis: two.sided
## 95 percent confidence interval:
##  -0.05486643  0.15801958
## sample estimates:
##    prop 1    prop 2 
## 0.8083333 0.7567568

prop.test(c(97,84),c(120,111),correct = FALSE)

Hypothesis Test for Difference of Proportions ExamplesLet's work some more problems together. 
70

Inference for a Sample MeanA potato chip manufacturer claims that there is on average 32 chips per bag for their brand. How can we tell if this is an accurate claim? 
71

Inference for a Sample Mean

A potato chip manufacturer claims that there is on average 32 chips per bag for their brand. How can we tell if this is an accurate claim?
One approach is to take a sample of, say 25 bags of chips for the particular band and compute the sample mean number of chips per bag.

Inference for a Sample Mean

A potato chip manufacturer claims that there is on average 32 chips per bag for their brand. How can we tell if this is an accurate claim?
One approach is to take a sample of, say 25 bags of chips for the particular band and compute the sample mean number of chips per bag.
This type of problem is inference for a sample mean.

Inference for a Sample Mean

A potato chip manufacturer claims that there is on average 32 chips per bag for their brand. How can we tell if this is an accurate claim?
One approach is to take a sample of, say 25 bags of chips for the particular band and compute the sample mean number of chips per bag.
This type of problem is inference for a sample mean.
In principle, we know the sampling distribution for the sample mean is very nearly normal, provided the conditions for the CLT holds.

Inference for a Sample Mean

A potato chip manufacturer claims that there is on average 32 chips per bag for their brand. How can we tell if this is an accurate claim?
One approach is to take a sample of, say 25 bags of chips for the particular band and compute the sample mean number of chips per bag.
This type of problem is inference for a sample mean.
In principle, we know the sampling distribution for the sample mean is very nearly normal, provided the conditions for the CLT holds.
One question is, what are the appropriate conditions to check to make sure the CLT holds for the sample mean?

Inference for a Sample Mean

A potato chip manufacturer claims that there is on average 32 chips per bag for their brand. How can we tell if this is an accurate claim?
One approach is to take a sample of, say 25 bags of chips for the particular band and compute the sample mean number of chips per bag.
This type of problem is inference for a sample mean.
In principle, we know the sampling distribution for the sample mean is very nearly normal, provided the conditions for the CLT holds.
One question is, what are the appropriate conditions to check to make sure the CLT holds for the sample mean?
When the conditions for the CLT for the sample mean hold, the expression for the standard error involves the population standard deviation ( $σ$ ) which we rarely know if practice. So, how do we estimate the standard error for the sample mean?

Inference for a Sample Mean

A potato chip manufacturer claims that there is on average 32 chips per bag for their brand. How can we tell if this is an accurate claim?
One approach is to take a sample of, say 25 bags of chips for the particular band and compute the sample mean number of chips per bag.
This type of problem is inference for a sample mean.
In principle, we know the sampling distribution for the sample mean is very nearly normal, provided the conditions for the CLT holds.
One question is, what are the appropriate conditions to check to make sure the CLT holds for the sample mean?
When the conditions for the CLT for the sample mean hold, the expression for the standard error involves the population standard deviation ( $σ$ ) which we rarely know if practice. So, how do we estimate the standard error for the sample mean?
We address these questions in the next few slides.

Inference for a Mean VideoYou are encouraged to watch this video on inference for a mean.

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Conditions for CLT for Sample MeanTwo conditions are required to apply the CLT for a sample mean ˉx¯x:
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Conditions for CLT for Sample Mean

Two conditions are required to apply the CLT for a sample mean $¯ x$ :

Independence: The sample observations must be independent. Simple random samples from a population are independent, as are data from a random process like rolling a die or tossing a coin.

Conditions for CLT for Sample Mean

Two conditions are required to apply the CLT for a sample mean $¯ x$ :

Independence: The sample observations must be independent. Simple random samples from a population are independent, as are data from a random process like rolling a die or tossing a coin. Normality: When a sample is small, we also require that the sample observations come from a normally distributed population. This condition can be relaxed for larger sample sizes.

Conditions for CLT for Sample Mean

Two conditions are required to apply the CLT for a sample mean $¯ x$ :

Independence: The sample observations must be independent. Simple random samples from a population are independent, as are data from a random process like rolling a die or tossing a coin. Normality: When a sample is small, we also require that the sample observations come from a normally distributed population. This condition can be relaxed for larger sample sizes.

The normality condition is vague but there are some approximate rules that work well in practice.

Practical Check for Normality n<30n<30: If the sample size nn is less than 30 and there are no clear outliers, then we can assume the data come from a nearly normal distribution. 
83

Practical Check for Normality

$n < 30$ : If the sample size $n$ is less than 30 and there are no clear outliers, then we can assume the data come from a nearly normal distribution.
$n \geq 30$ : If the sample size $n$ is at least 30 and there are no particularly exterme outliers, then we can assume the sampling distribution of $¯ x$ is nearly normal, even if the underlying distribution of individual observations is not.

Practical Check for Normality

$n < 30$ : If the sample size $n$ is less than 30 and there are no clear outliers, then we can assume the data come from a nearly normal distribution.
$n \geq 30$ : If the sample size $n$ is at least 30 and there are no particularly exterme outliers, then we can assume the sampling distribution of $¯ x$ is nearly normal, even if the underlying distribution of individual observations is not.
Let's consider example 7.1 from the textbook to illustrate these practical rules.

Estimating Standard ErrorWhen samples are independent and the normality condition is met, then the sampling distribution for the sample mean ˉx¯x is (very nearly) normal with mean μˉx=μμ¯x=μ and standard error SEˉx=σ√nSE¯x=σ√n, where μμ is the true population mean of the distribution from which the samples are taken and σσ is the true population from which the samples are taken. 
86

Estimating Standard Error

When samples are independent and the normality condition is met, then the sampling distribution for the sample mean $¯ x$ is (very nearly) normal with mean $μ_{¯ x} = μ$ and standard error $S E_{¯ x} = \frac{σ}{\sqrt{n}}$ , where $μ$ is the true population mean of the distribution from which the samples are taken and $σ$ is the true population from which the samples are taken.
In practice we do not know the true values of the population parameters $μ$ and $σ$ . But we can use the plug-in principle to obtain estimates:

$μ_{¯ x} \approx ¯ x, and S E_{¯ x} \approx \frac{s}{\sqrt{n}},$

where $s$ is the sample standard deviation.

T-score

If we take $n$ samples from a normal distribution $N (μ, σ)$ , then the quantity

$z = \frac{¯ x - μ}{\frac{σ}{\sqrt{n}}}$ will follow a standard normal distribution $N (0, 1)$ .

T-score

If we take $n$ samples from a normal distribution $N (μ, σ)$ , then the quantity

$z = \frac{¯ x - μ}{\frac{σ}{\sqrt{n}}}$ will follow a standard normal distribution $N (0, 1)$ .

On the other hand, the quantity

$t = \frac{¯ x - μ}{\frac{s}{\sqrt{n}}}$ will not quite be $N (0, 1)$ , especially if $n$ is small.

T-score

If we take $n$ samples from a normal distribution $N (μ, σ)$ , then the quantity

$z = \frac{¯ x - μ}{\frac{σ}{\sqrt{n}}}$ will follow a standard normal distribution $N (0, 1)$ .

On the other hand, the quantity

$t = \frac{¯ x - μ}{\frac{s}{\sqrt{n}}}$ will not quite be $N (0, 1)$ , especially if $n$ is small.

We call the quantity in the last formula a T-score.

T-Score VideoLet's watch this video on T-scores. 

91

t-distributionT-scores follow a t-distribution with degrees of freedom df=n−1df=n−1, where nn is the sample size. 
92

t-distribution

T-scores follow a t-distribution with degrees of freedom $df = n - 1$ , where $n$ is the sample size.
We will use t-distributions for inference, that is, for obtaining confidence intervals and hypothesis testing.

t-distribution

T-scores follow a t-distribution with degrees of freedom $df = n - 1$ , where $n$ is the sample size.
We will use t-distributions for inference, that is, for obtaining confidence intervals and hypothesis testing.
Let's get a feel for t-distributions.

Plotting t-Distributions

We can easily plot density curves for t-distributions.

gf_dist("t",df=19)

Degrees of Freedom Parameter

This plot shows three different t density functions for three different degrees of freedom:

A Computational Experiment

This plot shows a histogram of T-score values obtained by computing the sample mean with $n = 8$ from $N (0, 1)$ . We overlay a density curve for both $N (0, 1)$ and a t-distribution with $df = 7$ :

Area Under a t-Distribution

We can compute areas under t-distribution density curves in the same way we did for areas under normal distribution density curves. For example, the area

is computed by

pt(-1.0,df=5)

## [1] 0.1816087

Middle 95% Under a t DensityHow do we find the middle 95% of area under a t-distribution? This is an important question because it relates to construting a 95% confidence interval for the sample mean. 
99

Middle 95% Under a t Density

How do we find the middle 95% of area under a t-distribution? This is an important question because it relates to construting a 95% confidence interval for the sample mean.
Suppose we have a t-distribution with degrees of freedom $df = 10$ , then to find the value $t^{*}$ so that 95% of the area under the density curve lies between $- t^{*}$ and $t^{*}$ , we use the qt command, for example,

## [1] 2.228139

(t_ast <- -qt(0.05/2,df=10))

100

Middle 95% Under a t Density

How do we find the middle 95% of area under a t-distribution? This is an important question because it relates to construting a 95% confidence interval for the sample mean.
Suppose we have a t-distribution with degrees of freedom $df = 10$ , then to find the value $t^{*}$ so that 95% of the area under the density curve lies between $- t^{*}$ and $t^{*}$ , we use the qt command, for example,

## [1] 2.228139

(t_ast <- -qt(0.05/2,df=10))

We can check our answer:

## [1] 0.95

pt(t_ast,df=10) - pt(-t_ast,df=10)

101

Middle 95% Under a t Density

How do we find the middle 95% of area under a t-distribution? This is an important question because it relates to construting a 95% confidence interval for the sample mean.
Suppose we have a t-distribution with degrees of freedom $df = 10$ , then to find the value $t^{*}$ so that 95% of the area under the density curve lies between $- t^{*}$ and $t^{*}$ , we use the qt command, for example,

## [1] 2.228139

(t_ast <- -qt(0.05/2,df=10))

We can check our answer:

## [1] 0.95

pt(t_ast,df=10) - pt(-t_ast,df=10)

Note that you always have to specify the appropriate value for the degrees of freedom df. The appropriate degrees of freedom is given by $df = n - 1$ , where $n$ is the sample size.

102

One-Sample t CI

Based on a sample of $n$ independent and nearly normal observations, a confidence interval for the population mean is $¯ x \pm t_{df}^{*} \times \frac{s}{\sqrt{n}},$

where $n$ is the sample size, $¯ x$ is the sample mean, $s$ is the sample standard deviation.

103

One-Sample t CI

Based on a sample of $n$ independent and nearly normal observations, a confidence interval for the population mean is $¯ x \pm t_{df}^{*} \times \frac{s}{\sqrt{n}},$

where $n$ is the sample size, $¯ x$ is the sample mean, $s$ is the sample standard deviation.

We determine the appropriate value for $t_{df}^{*}$ with the R command

qt((1.0-confidence_level)/2,df=n-1).

104

One-Sample Mean CI Example

Suppose we take a random sample of 13 observations from a normally distributed population and determine the sample mean is 8 with a sample standard deviation of 2.5. Then to find a 90% confidence interval, we would do the following:

## [1] 6.764206 9.235794

n <- 13
x_bar <- 8
s <- 2.5
SE <- s/sqrt(n)
t_ast <- -qt((1.0-0.9)/2,df=n-1)
(CI <- x_bar + t_ast * c(-1,1)*SE)

A 95% CI would be

## [1] 6.489265 9.510735

t_ast <- -qt((1.0-0.95)/2,df=n-1)
(CI <- x_bar + t_ast * c(-1,1)*SE)

105

CI for Single Mean Summary

Once you have determined a one-mean confidence interval would be helpful for an application, there are four steps to constructing the interval:

106

CI for Single Mean Summary

Once you have determined a one-mean confidence interval would be helpful for an application, there are four steps to constructing the interval:

Identify $n$ , $¯ x$ , and $s$ , and determine what confidence level you wish to use.

107

CI for Single Mean Summary

Once you have determined a one-mean confidence interval would be helpful for an application, there are four steps to constructing the interval:

Identify $n$ , $¯ x$ , and $s$ , and determine what confidence level you wish to use.

Verify the conditions to ensure $¯ x$ is nearly normal.

108

CI for Single Mean Summary

Once you have determined a one-mean confidence interval would be helpful for an application, there are four steps to constructing the interval:

Identify $n$ , $¯ x$ , and $s$ , and determine what confidence level you wish to use.

Verify the conditions to ensure $¯ x$ is nearly normal.
If the conditions hold, approximate $S E$ by $\frac{s}{\sqrt{n}}$ , find $t_{df}^{*}$ , and construct the interval.

109

CI for Single Mean Summary

Once you have determined a one-mean confidence interval would be helpful for an application, there are four steps to constructing the interval:

Identify $n$ , $¯ x$ , and $s$ , and determine what confidence level you wish to use.

Verify the conditions to ensure $¯ x$ is nearly normal.
If the conditions hold, approximate $S E$ by $\frac{s}{\sqrt{n}}$ , find $t_{df}^{*}$ , and construct the interval.
Interpret the confidence interval in the context of the problem.

110

One-Mean Hypothesis TestingThe null hypothesis for a one-proportion test is typically stated as H0:μ=μ0H0:μ=μ0 where μ0μ0 is the null value for the population mean. 
111

One-Mean Hypothesis Testing

The null hypothesis for a one-proportion test is typically stated as $H_{0} : μ = μ_{0}$ where $μ_{0}$ is the null value for the population mean.
The corresponding alternative hypothesis is then one of

112

One-Mean Hypothesis Testing

The null hypothesis for a one-proportion test is typically stated as $H_{0} : μ = μ_{0}$ where $μ_{0}$ is the null value for the population mean.
The corresponding alternative hypothesis is then one of
- $H_{A} : μ \neq μ_{0}$ (two-sided),

113

One-Mean Hypothesis Testing

The null hypothesis for a one-proportion test is typically stated as $H_{0} : μ = μ_{0}$ where $μ_{0}$ is the null value for the population mean.
The corresponding alternative hypothesis is then one of
- $H_{A} : μ \neq μ_{0}$ (two-sided),
- $H_{A} : μ < μ_{0}$ (one-sided less than), or

114

One-Mean Hypothesis Testing

The null hypothesis for a one-proportion test is typically stated as $H_{0} : μ = μ_{0}$ where $μ_{0}$ is the null value for the population mean.
The corresponding alternative hypothesis is then one of
- $H_{A} : μ \neq μ_{0}$ (two-sided),
- $H_{A} : μ < μ_{0}$ (one-sided less than), or
- $H_{A} : μ > μ_{0}$ (one-sided greater than)

115

One-Mean Hypothesis Test ProcedureOnce you have determined a one-mean hypothesis test is the correct procedure, there are four steps to completing the test:
116

One-Mean Hypothesis Test Procedure

Once you have determined a one-mean hypothesis test is the correct procedure, there are four steps to completing the test:
- Identify the parameter of interest, list out hypotheses, identify the significance level, and identify $n$ , $¯ x$ , and $s$ .

117

One-Mean Hypothesis Test Procedure

Once you have determined a one-mean hypothesis test is the correct procedure, there are four steps to completing the test:
- Identify the parameter of interest, list out hypotheses, identify the significance level, and identify $n$ , $¯ x$ , and $s$ .
- Verify conditions to ensure $¯ x$ is nearly normal.

118

One-Mean Hypothesis Test Procedure

Once you have determined a one-mean hypothesis test is the correct procedure, there are four steps to completing the test:
- Identify the parameter of interest, list out hypotheses, identify the significance level, and identify $n$ , $¯ x$ , and $s$ .
- Verify conditions to ensure $¯ x$ is nearly normal.
- If the conditions hold, approximate $S E$ by $\frac{s}{\sqrt{n}}$ , compute the T-score using
$T = \frac{¯ x - μ_{0}}{\frac{s}{\sqrt{n}}},$ and compute the p-value.

119

One-Mean Hypothesis Test Procedure

Once you have determined a one-mean hypothesis test is the correct procedure, there are four steps to completing the test:
- Identify the parameter of interest, list out hypotheses, identify the significance level, and identify $n$ , $¯ x$ , and $s$ .
- Verify conditions to ensure $¯ x$ is nearly normal.
- If the conditions hold, approximate $S E$ by $\frac{s}{\sqrt{n}}$ , compute the T-score using
$T = \frac{¯ x - μ_{0}}{\frac{s}{\sqrt{n}}},$ and compute the p-value.
- Evaluate the hypothesis test by comparing the p-value to the significance level $α$ , and provide a conclusion in the context of the problem.

120

A Simple Example

We want to know if the following data comes from a normal distribution with mean 0:

##  [1] -0.36047565 -0.03017749  1.75870831  0.27050839  0.32928774  1.91506499
##  [7]  0.66091621 -1.06506123 -0.48685285 -0.24566197

We can apply a hypothesis test corresponding to $H_{0} : μ = 0.0$ , versus $H_{A} : μ \neq 0$ as follows

## [1] 0.9105232

## [1] 0.3862831

n <- length(x); x_bar <- mean(x); s <- sd(x)
alpha <- 0.05
SE <- s/sqrt(n)
mu_0 <- 0.0
(T <- (x_bar - mu_0)/SE)
(p_value <- 2*(1-pt(T,df=n-1)))

121

R Command for One-Sample t-testThere is a built-in R command, t.test that will conduct a hypothesis test for a sinlg emean for us. 
122

R Command for One-Sample t-test

There is a built-in R command, t.test that will conduct a hypothesis test for a sinlg emean for us.
For example, we can solve our previous problem using

## 
##     One Sample t-test
## 
## data:  x
## t = 0.91052, df = 9, p-value = 0.3863
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  -0.4076704  0.9569217
## sample estimates:
## mean of x 
## 0.2746256

t.test(x,mu=0.0)

123

R Command for One-Sample t-test

There is a built-in R command, t.test that will conduct a hypothesis test for a sinlg emean for us.
For example, we can solve our previous problem using

## 
##     One Sample t-test
## 
## data:  x
## t = 0.91052, df = 9, p-value = 0.3863
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  -0.4076704  0.9569217
## sample estimates:
## mean of x 
## 0.2746256

t.test(x,mu=0.0)

Let's work some examples together.

124

Paired Data

Two sets of observations are paired if each observation in one set has a special correspondence or connection with exactly one observation in the other data set.

125

Paired Data

Two sets of observations are paired if each observation in one set has a special correspondence or connection with exactly one observation in the other data set.

Common examples of paired data correspond to "before" and "after" trials.

126

Paired Data

Two sets of observations are paired if each observation in one set has a special correspondence or connection with exactly one observation in the other data set.

Common examples of paired data correspond to "before" and "after" trials.

For example, does a particular study technique work well for increasing one's exam score? To test this, we can ask 25 people to take an exam and record their scores, then we can ask those same people to try the study technique before taking another similar exam.

127

Paired Data

Two sets of observations are paired if each observation in one set has a special correspondence or connection with exactly one observation in the other data set.

Common examples of paired data correspond to "before" and "after" trials.

For example, does a particular study technique work well for increasing one's exam score? To test this, we can ask 25 people to take an exam and record their scores, then we can ask those same people to try the study technique before taking another similar exam.

As another example, suppose it is claimed that among the general population of adults in the US, the average length of the left foot is longer than the average length of the right foot. To test this, we can select 32 people, record the measurement of the left foot of everyone in one column, then record the measurement of the right foot of everyone in a second column. We must make sure that each row of the resulting data frame corresponds with only one person.

128

Paired Data Example

Perhaps the data from our last example looks as follows:

## # A tibble: 6 × 3
##   left_foot right_foot foot_diff
##       <dbl>      <dbl>     <dbl>
## 1      7.83       8.27    -0.437
## 2      7.93       8.26    -0.332
## 3      8.47       8.25     0.221
## 4      8.02       8.21    -0.185
## 5      8.04       8.17    -0.127
## 6      8.51       7.98     0.533

129

Paired Data Example

Perhaps the data from our last example looks as follows:

## # A tibble: 6 × 3
##   left_foot right_foot foot_diff
##       <dbl>      <dbl>     <dbl>
## 1      7.83       8.27    -0.437
## 2      7.93       8.26    -0.332
## 3      8.47       8.25     0.221
## 4      8.02       8.21    -0.185
## 5      8.04       8.17    -0.127
## 6      8.51       7.98     0.533

Notice that we have added a column that is the difference of the left foot measurement minus the right foot measurement.

130

Hypothesis Test for Paired DataHow can we set up a hypothesis testing framework for the foot measurement question? 
131

Hypothesis Test for Paired Data

How can we set up a hypothesis testing framework for the foot measurement question?
Basically, we can apply statistical inference to the difference column in the data.

132

Hypothesis Test for Paired Data

How can we set up a hypothesis testing framework for the foot measurement question?
Basically, we can apply statistical inference to the difference column in the data.
The typical null hypothesis for paired data is that the average difference between the measurements is 0. We write this as

$H_{0} : μ_{d} = 0$

133

Paired Hypothesis Test ProcedureOnce you have determined a paired hypothesis test is the correct procedure, there are four steps to completing the test:
134

Paired Hypothesis Test Procedure

Once you have determined a paired hypothesis test is the correct procedure, there are four steps to completing the test:
- Determine the significance level, the sample size $n$ , the mean of the differences $¯ d$ , and the corresponding standard deviation $s_{d}$ .

135

Paired Hypothesis Test Procedure

Once you have determined a paired hypothesis test is the correct procedure, there are four steps to completing the test:
- Determine the significance level, the sample size $n$ , the mean of the differences $¯ d$ , and the corresponding standard deviation $s_{d}$ .
- Verify the conditions to ensure that $¯ d$ is nearly normal.

136

Paired Hypothesis Test Procedure

Once you have determined a paired hypothesis test is the correct procedure, there are four steps to completing the test:
- Determine the significance level, the sample size $n$ , the mean of the differences $¯ d$ , and the corresponding standard deviation $s_{d}$ .
- Verify the conditions to ensure that $¯ d$ is nearly normal.
- If the conditions hold, approximate $S E$ by $\frac{s_{d}}{\sqrt{n}}$ , compute the T-score using
$T = \frac{¯ d}{\frac{s_{d}}{\sqrt{n}}}$ and compute the p-value.

137

Paired Hypothesis Test Procedure

Once you have determined a paired hypothesis test is the correct procedure, there are four steps to completing the test:
- Determine the significance level, the sample size $n$ , the mean of the differences $¯ d$ , and the corresponding standard deviation $s_{d}$ .
- Verify the conditions to ensure that $¯ d$ is nearly normal.
- If the conditions hold, approximate $S E$ by $\frac{s_{d}}{\sqrt{n}}$ , compute the T-score using
$T = \frac{¯ d}{\frac{s_{d}}{\sqrt{n}}}$ and compute the p-value.
- Evaluate the hypothesis test by comparing the p-value to the significance level $α$ , and provide a conclusion in the context of the problem.

138

First ExampleConsider our foot measurement data.The sample size is n=32n=32 and a boxplot shows that there are no extreme outliers. Now we compute the necessary quantities:
Result
Code
## [1] -0.6907872
## [1] 0.4948396
n <- 32
d_bar <- mean(foot_df$foot_diff)
s_d <- sd(foot_df$foot_diff)
(t_val <- d_bar/(s_d/sqrt(n)))
(p_val <- 2*pt(t_val,df=n-1))
139

First ExampleConsider our foot measurement data.The sample size is n=32n=32 and a boxplot shows that there are no extreme outliers. Now we compute the necessary quantities:
Result
Code
## [1] -0.6907872
## [1] 0.4948396
n <- 32
d_bar <- mean(foot_df$foot_diff)
s_d <- sd(foot_df$foot_diff)
(t_val <- d_bar/(s_d/sqrt(n)))
(p_val <- 2*pt(t_val,df=n-1))
Here we fail to reject the null hypothesis at the 0.05 significance level. That is, the data does not provide sufficient evidence for rejecting the null hypothesis that there is no significant difference in the length of the left foot versus the right foot.
140

R Command for Paired Hypothesis TestAgain, we can use the t.test function. However, now we must use two sets of data and add the paired=TRUE argument. 
141

R Command for Paired Hypothesis Test

Again, we can use the t.test function. However, now we must use two sets of data and add the paired=TRUE argument.
For example, to test the hypothesis for the foot data, one would use

## 
##     Paired t-test
## 
## data:  foot_df$left_foot and foot_df$right_foot
## t = -0.69079, df = 31, p-value = 0.4948
## alternative hypothesis: true mean difference is not equal to 0
## 95 percent confidence interval:
##  -0.18623480  0.09199711
## sample estimates:
## mean difference 
##     -0.04711885

t.test(foot_df$left_foot,foot_df$right_foot,paired = TRUE)

142

R Command for Paired Hypothesis Test

Again, we can use the t.test function. However, now we must use two sets of data and add the paired=TRUE argument.
For example, to test the hypothesis for the foot data, one would use

## 
##     Paired t-test
## 
## data:  foot_df$left_foot and foot_df$right_foot
## t = -0.69079, df = 31, p-value = 0.4948
## alternative hypothesis: true mean difference is not equal to 0
## 95 percent confidence interval:
##  -0.18623480  0.09199711
## sample estimates:
## mean difference 
##     -0.04711885

t.test(foot_df$left_foot,foot_df$right_foot,paired = TRUE)

Let's work some more examples together.

143

Difference of Two MeansInference for paired data compares two different (but related) measurements on the same population. For example, we might want to study the difference in how individuals sleep before and after consuming a large amount of caffeine. 
144

Difference of Two Means

Inference for paired data compares two different (but related) measurements on the same population. For example, we might want to study the difference in how individuals sleep before and after consuming a large amount of caffeine.
On the other hand, inference for a difference of two means compares the same measurement on two difference populations. For example, we may want to study any difference between how caffeine affects the sleep of individuals with high blood pressure compared to those that do not have high blood pressure.

145

Difference of Two Means

Inference for paired data compares two different (but related) measurements on the same population. For example, we might want to study the difference in how individuals sleep before and after consuming a large amount of caffeine.
On the other hand, inference for a difference of two means compares the same measurement on two difference populations. For example, we may want to study any difference between how caffeine affects the sleep of individuals with high blood pressure compared to those that do not have high blood pressure.
We can still use a t-distribution for inference for a difference of two means but we must compute the two-sample means and standard deviations separately for estimating standard error.

146

Difference of Two Means

Inference for paired data compares two different (but related) measurements on the same population. For example, we might want to study the difference in how individuals sleep before and after consuming a large amount of caffeine.
On the other hand, inference for a difference of two means compares the same measurement on two difference populations. For example, we may want to study any difference between how caffeine affects the sleep of individuals with high blood pressure compared to those that do not have high blood pressure.
We can still use a t-distribution for inference for a difference of two means but we must compute the two-sample means and standard deviations separately for estimating standard error.

We proceed with the details.

147

Confidence Intervals for a Difference in MeansThe t-distribution can be used for inference when working with the standardized difference of two means if
148

Confidence Intervals for a Difference in Means

The t-distribution can be used for inference when working with the standardized difference of two means if
- The data are independent within and between the two groups, e.g., the data come from independent random samples or from a randomized experiment.

149

Confidence Intervals for a Difference in Means

The t-distribution can be used for inference when working with the standardized difference of two means if
- The data are independent within and between the two groups, e.g., the data come from independent random samples or from a randomized experiment.
- We check the outliers rules of thumb for each group separately.

150

Confidence Intervals for a Difference in Means

The t-distribution can be used for inference when working with the standardized difference of two means if
- The data are independent within and between the two groups, e.g., the data come from independent random samples or from a randomized experiment.
- We check the outliers rules of thumb for each group separately.
The standard error may be computed as

$S E = \sqrt{\frac{σ_{1}^{2}}{n_{1}} + \frac{σ_{2}^{2}}{n_{2}}}$

151

Hypothesis Tests for Difference of MeansHypothesis tests for a difference of two means works in a very similar fashion to what we have seen before. 
152

Hypothesis Tests for Difference of Means

Hypothesis tests for a difference of two means works in a very similar fashion to what we have seen before.
To conduct a test for a difference of two means "by hand", use the smaller of the two degrees of freedom.

153

Hypothesis Tests for Difference of Means

Hypothesis tests for a difference of two means works in a very similar fashion to what we have seen before.
To conduct a test for a difference of two means "by hand", use the smaller of the two degrees of freedom.
Use the t.test function without the paired = TRUE argument.

154

Examples of Inference for Difference of MeansLet's look at some examples and work some problems together. 
155

Statistical PowerOur next topic is statistical power, see the following video to get an introduction. 

156

Notes157

Notes158

Notes159

MATH 204 Introduction to Statistics
Lecture 12: Further Inference
JMG
1

Goals for LectureIn this lecture, we introduce statistical inference for
2

Goals for Lecture

In this lecture, we introduce statistical inference for
- a single proportion (6.1)

Goals for Lecture

In this lecture, we introduce statistical inference for
- a single proportion (6.1)
- a difference of two proportions (6.2)

Goals for Lecture

In this lecture, we introduce statistical inference for
- a single proportion (6.1)
- a difference of two proportions (6.2)
- one-sample means (7.1)

Goals for Lecture

In this lecture, we introduce statistical inference for
- a single proportion (6.1)
- a difference of two proportions (6.2)
- one-sample means (7.1)
- paired data (7.3)

Goals for Lecture

In this lecture, we introduce statistical inference for
- a single proportion (6.1)
- a difference of two proportions (6.2)
- one-sample means (7.1)
- paired data (7.3)
- a difference of two means (7.3)

Goals for Lecture

In this lecture, we introduce statistical inference for
- a single proportion (6.1)
- a difference of two proportions (6.2)
- one-sample means (7.1)
- paired data (7.3)
- a difference of two means (7.3)
After we discuss these topics along with power calculations and simple ordinary least squares regression (7.4 and Chapter 8), we will return to discuss goodness of fit (6.3) and testing for independence (6.4).

Learning ObjectivesAfter this lecture, you should be able to conduct and apply point estimates, interval estimates, and hypothesis tests for
9

Learning Objectives

After this lecture, you should be able to conduct and apply point estimates, interval estimates, and hypothesis tests for
- proportions, difference of proportions, one sample means, paired data, and difference of two means.

Learning Objectives

After this lecture, you should be able to conduct and apply point estimates, interval estimates, and hypothesis tests for
- proportions, difference of proportions, one sample means, paired data, and difference of two means.
This includes knowing when to use which type of hypothesis test, and

Learning Objectives

After this lecture, you should be able to conduct and apply point estimates, interval estimates, and hypothesis tests for
- proportions, difference of proportions, one sample means, paired data, and difference of two means.
This includes knowing when to use which type of hypothesis test, and
what is the appropriate R command(s) to use.

Learning Objectives

After this lecture, you should be able to conduct and apply point estimates, interval estimates, and hypothesis tests for
- proportions, difference of proportions, one sample means, paired data, and difference of two means.
This includes knowing when to use which type of hypothesis test, and
what is the appropriate R command(s) to use.
Many of our interval estimates and hypothesis tests will rely on a central limit theorem. We quickly review these.

Central Limit Theorems

Recall our CLT for sample proportions:

When observations are independent and sample size is sufficiently large, then the sampling distribution for the sample proportion $^p$ is approximately normal with mean $μ_{^p} = p$ and standard error $S E_{^p} = \sqrt{\frac{p (1 - p)}{n}}$ .

Central Limit Theorems

Recall our CLT for sample proportions:

When observations are independent and sample size is sufficiently large, then the sampling distribution for the sample proportion $^p$ is approximately normal with mean $μ_{^p} = p$ and standard error $S E_{^p} = \sqrt{\frac{p (1 - p)}{n}}$ .

We also have a CLT for the sample mean:

When we collect samples of sufficiently large size $n$ from a population with mean $μ$ and standard deviation $σ$ , then the sampling distribution for the sample mean $¯ x$ is approximately normal with mean $μ_{¯ x} = μ$ and standard error $S E_{¯ x} = \frac{σ}{\sqrt{n}}$ .

Central Limit Theorems

Recall our CLT for sample proportions:

When observations are independent and sample size is sufficiently large, then the sampling distribution for the sample proportion $^p$ is approximately normal with mean $μ_{^p} = p$ and standard error $S E_{^p} = \sqrt{\frac{p (1 - p)}{n}}$ .

We also have a CLT for the sample mean:

When we collect samples of sufficiently large size $n$ from a population with mean $μ$ and standard deviation $σ$ , then the sampling distribution for the sample mean $¯ x$ is approximately normal with mean $μ_{¯ x} = μ$ and standard error $S E_{¯ x} = \frac{σ}{\sqrt{n}}$ .

Later we will return to discuss precisely what we mean by "samples of sufficiently large size."

Central Limit Theorems

Recall our CLT for sample proportions:

When observations are independent and sample size is sufficiently large, then the sampling distribution for the sample proportion $^p$ is approximately normal with mean $μ_{^p} = p$ and standard error $S E_{^p} = \sqrt{\frac{p (1 - p)}{n}}$ .

We also have a CLT for the sample mean:

When we collect samples of sufficiently large size $n$ from a population with mean $μ$ and standard deviation $σ$ , then the sampling distribution for the sample mean $¯ x$ is approximately normal with mean $μ_{¯ x} = μ$ and standard error $S E_{¯ x} = \frac{σ}{\sqrt{n}}$ .

Later we will return to discuss precisely what we mean by "samples of sufficiently large size."

We use sampling distributions for obtaining confidence intervals and conducting hypothesis tests.

Videos for CLTCentral limit theorems are reviewed in the videos included on the next few slides.
18

CLT Video 1

19

CLT Video 2

20

CLT Video 3

21

CI for a ProportionOnce you've determined a one-proportion CI would be helpful for an application, there are four steps to constructing the interval:
22

CI for a Proportion

Once you've determined a one-proportion CI would be helpful for an application, there are four steps to constructing the interval:
- Identify the point estimate $^p$ and the sample size $n$ , and determine what confidence level you want.

CI for a Proportion

Once you've determined a one-proportion CI would be helpful for an application, there are four steps to constructing the interval:
- Identify the point estimate $^p$ and the sample size $n$ , and determine what confidence level you want.
- Verify the conditions to ensure $^p$ is nearly normal, that is, check that $n^p \geq 10$ and $n (1 -^p) \geq 10$ .

CI for a Proportion

Once you've determined a one-proportion CI would be helpful for an application, there are four steps to constructing the interval:
- Identify the point estimate $^p$ and the sample size $n$ , and determine what confidence level you want.
- Verify the conditions to ensure $^p$ is nearly normal, that is, check that $n^p \geq 10$ and $n (1 -^p) \geq 10$ .
- If the conditions hold, estimate the standard error $S E$ using $^p$ , find the appropriate $z^{*}$ , and compute $^p \pm z^{*} \times S E$ .

CI for a Proportion

Once you've determined a one-proportion CI would be helpful for an application, there are four steps to constructing the interval:
- Identify the point estimate $^p$ and the sample size $n$ , and determine what confidence level you want.
- Verify the conditions to ensure $^p$ is nearly normal, that is, check that $n^p \geq 10$ and $n (1 -^p) \geq 10$ .
- If the conditions hold, estimate the standard error $S E$ using $^p$ , find the appropriate $z^{*}$ , and compute $^p \pm z^{*} \times S E$ .
- Interpret the CI in the context of the problem.

Proportion CI ExampleConsider the following problem:
27

Proportion CI Example

Consider the following problem:

Is parking a problem on campus? A randomly selected group of 89 faculty and staff are asked whether they are satisfied with campus parking or not. Of the 89 individuals surveyed, 23 indicated that they are satisfied. What is the proportion $p$ of faculty and staff that are satisfied with campus parking?

Proportion CI Example

Consider the following problem:

Is parking a problem on campus? A randomly selected group of 89 faculty and staff are asked whether they are satisfied with campus parking or not. Of the 89 individuals surveyed, 23 indicated that they are satisfied. What is the proportion $p$ of faculty and staff that are satisfied with campus parking?

Explain why this problem is suitable for a one-proportion CI. Is this CLT applicable? If so, obtain a 95% CI for a point estimate.

Proportion CI VideoConfidence intervals for a proportion are reviewed in this video:

30

Hypothesis Test for a ProportionOnce you've determined a one-proportion hypothesis test is the correct test for a problem, there are four steps to completing the test:
31

Hypothesis Test for a Proportion

Once you've determined a one-proportion hypothesis test is the correct test for a problem, there are four steps to completing the test:
- Identify the parameter of interest, list hypotheses, identify the significance level, and identify $^p$ and $n$ .

Hypothesis Test for a Proportion

Once you've determined a one-proportion hypothesis test is the correct test for a problem, there are four steps to completing the test:
- Identify the parameter of interest, list hypotheses, identify the significance level, and identify $^p$ and $n$ .
- Verify conditions to ensure $^p$ is nearly normal under $H_{0}$ . For one-proportion hypothesis tests, use the null value $p_{0}$ to check the conditions $n p_{0} \geq 10$ and $n (1 - p_{0}) \geq 10$ .

Hypothesis Test for a Proportion

Once you've determined a one-proportion hypothesis test is the correct test for a problem, there are four steps to completing the test:
- Identify the parameter of interest, list hypotheses, identify the significance level, and identify $^p$ and $n$ .
- Verify conditions to ensure $^p$ is nearly normal under $H_{0}$ . For one-proportion hypothesis tests, use the null value $p_{0}$ to check the conditions $n p_{0} \geq 10$ and $n (1 - p_{0}) \geq 10$ .
- If the conditions hold, compute the standard error, again using $p_{0}$ , compute the $Z$ -score, and identify the p-value.

Hypothesis Test for a Proportion

Once you've determined a one-proportion hypothesis test is the correct test for a problem, there are four steps to completing the test:
- Identify the parameter of interest, list hypotheses, identify the significance level, and identify $^p$ and $n$ .
- Verify conditions to ensure $^p$ is nearly normal under $H_{0}$ . For one-proportion hypothesis tests, use the null value $p_{0}$ to check the conditions $n p_{0} \geq 10$ and $n (1 - p_{0}) \geq 10$ .
- If the conditions hold, compute the standard error, again using $p_{0}$ , compute the $Z$ -score, and identify the p-value.
- Evaluate the hypothesis test by comparing the p-value to the significance level $α$ , and provide a conclusion in the context of the problem.

Hypotheses for One Proportion TestsThe null hypothesis for a one-proportion test is typically stated as H0:p=p0H0:p=p0 where p0p0 is a hypothetical value for pp. 
36

Hypotheses for One Proportion Tests

The null hypothesis for a one-proportion test is typically stated as $H_{0} : p = p_{0}$ where $p_{0}$ is a hypothetical value for $p$ .
The corresponding alternative hypothesis is then one of

Hypotheses for One Proportion Tests

The null hypothesis for a one-proportion test is typically stated as $H_{0} : p = p_{0}$ where $p_{0}$ is a hypothetical value for $p$ .
The corresponding alternative hypothesis is then one of
- $H_{A} : p \neq p_{0}$ (two-sided),

Hypotheses for One Proportion Tests

The null hypothesis for a one-proportion test is typically stated as $H_{0} : p = p_{0}$ where $p_{0}$ is a hypothetical value for $p$ .
The corresponding alternative hypothesis is then one of
- $H_{A} : p \neq p_{0}$ (two-sided),
- $H_{A} : p < p_{0}$ (one-sided less than), or

Hypotheses for One Proportion Tests

The null hypothesis for a one-proportion test is typically stated as $H_{0} : p = p_{0}$ where $p_{0}$ is a hypothetical value for $p$ .
The corresponding alternative hypothesis is then one of
- $H_{A} : p \neq p_{0}$ (two-sided),
- $H_{A} : p < p_{0}$ (one-sided less than), or
- $H_{A} : p > p_{0}$ (one-sided greater than)

Proportion Hypothesis Test Example

Is parking a problem on campus? A randomly selected group of 89 faculty and staff are asked whether they are satisfied with campus parking or not. Of the 89 individuals surveyed, 23 indicated that they are satisfied.

Proportion Hypothesis Test Example

Is parking a problem on campus? A randomly selected group of 89 faculty and staff are asked whether they are satisfied with campus parking or not. Of the 89 individuals surveyed, 23 indicated that they are satisfied.

Let's study this problem using a hypothesis testing framework.

Proportion Hypothesis Test Example

Is parking a problem on campus? A randomly selected group of 89 faculty and staff are asked whether they are satisfied with campus parking or not. Of the 89 individuals surveyed, 23 indicated that they are satisfied.

Let's study this problem using a hypothesis testing framework.

Here is relevant R code:

## [1] 44.5 44.5

## [1] -4.557991

## [1] 5.164528e-06

p_hat <- 23/89
n <- 89
p0 <- 0.5
(c(n*p0,n*(1-p0))) # success-failure condition
SE <- sqrt((p0*(1-p0))/n)
(z_val <- (p_hat - p0)/SE)
(p_value <- 2*(pnorm(z_val)))

R Command(s) for Proportion Hypothesis TestIn cases where the CLT applies, we can use a normal distribution to directly compute p-values with the pnorm function. 
44

R Command(s) for Proportion Hypothesis Test

In cases where the CLT applies, we can use a normal distribution to directly compute p-values with the pnorm function.
Alternatively, R has a built-in function prop.test that automates one-proportion hypothesis testing.

R Command(s) for Proportion Hypothesis Test

In cases where the CLT applies, we can use a normal distribution to directly compute p-values with the pnorm function.
Alternatively, R has a built-in function prop.test that automates one-proportion hypothesis testing.
So, in our parking problem example we would use

## 
##     1-sample proportions test without continuity correction
## 
## data:  23 out of 89, null probability 0.5
## X-squared = 20.775, df = 1, p-value = 5.165e-06
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
##  0.1788154 0.3580294
## sample estimates:
##        p 
## 0.258427

prop.test(23,89,p=0.5,correct = FALSE)

Problems to PracticeLet's take a minute to do some practice problems. 
47

Problems to Practice

Let's take a minute to do some practice problems.
Try problems 6.7 and 6.13 from the textbook.

Problems Involving Difference of Proportions

Consider the following problem:

An online poll on January 10, 2017 reported that 97 out of 120 people in Virginia between the ages of 18 and 29 believe that marijuana should be legal, while 84 out of 111 who are 30 and over held this belief. Is there a difference between the proportion of young people who favor marijuana leagalization as compared to people who are older?

Problems Involving Difference of Proportions

Consider the following problem:

An online poll on January 10, 2017 reported that 97 out of 120 people in Virginia between the ages of 18 and 29 believe that marijuana should be legal, while 84 out of 111 who are 30 and over held this belief. Is there a difference between the proportion of young people who favor marijuana leagalization as compared to people who are older?

Think about how this problem is different than problems for a single proportion. The basic point here is that there are two groups, and we want to compare proportions across the two groups.

Problems Involving Difference of Proportions

Consider the following problem:

An online poll on January 10, 2017 reported that 97 out of 120 people in Virginia between the ages of 18 and 29 believe that marijuana should be legal, while 84 out of 111 who are 30 and over held this belief. Is there a difference between the proportion of young people who favor marijuana leagalization as compared to people who are older?

Think about how this problem is different than problems for a single proportion. The basic point here is that there are two groups, and we want to compare proportions across the two groups.

Data for the problem stated above might look as follows (only the first few rows are shown).

## # A tibble: 6 × 2
##   support_legalize age_group
##   <chr>            <fct>    
## 1 Yes              >30      
## 2 No               18-29    
## 3 Yes              >30      
## 4 Yes              >30      
## 5 Yes              >30      
## 6 Yes              18-29

Plotting Grouped Proportion Data

We can used grouped bar plots or a mosaic plot to visualize data related to grouped proportions:

Sampling Distribution for Difference of Proportions

The difference ${^p}_{1} - {^p}_{2}$ can be modeled using a normal distribution when

The data are independent within and between the two groups. Generally this is satisfied if the data come from two independent random samples or if the data come from a randomized experiment.
The success-failure condition holds for both groups, where we check successes and failures in each group separately.

Sampling Distribution for Difference of Proportions

The difference ${^p}_{1} - {^p}_{2}$ can be modeled using a normal distribution when

The data are independent within and between the two groups. Generally this is satisfied if the data come from two independent random samples or if the data come from a randomized experiment.
The success-failure condition holds for both groups, where we check successes and failures in each group separately.

When these conditions are satisfied, the standard error of ${^p}_{1} - {^p}_{2}$ is

$S E = \sqrt{\frac{p_{1} (1 - p_{1})}{n_{1}} + \frac{p_{2} (1 - p_{2})}{n_{2}}},$ where $p_{1}$ and $p_{2}$ represent the population proportions, and $n_{1}$ and $n_{2}$ represent the sample sizes.

CI for Difference of Proportions

To construct a CI for a difference of proportion, we can apply the formulas

$S E = \sqrt{\frac{p_{1} (1 - p_{1})}{n_{1}} + \frac{p_{2} (1 - p_{2})}{n_{2}}}, C I = point estimate \pm z^{*} \times S E$

CI for Difference of Proportions

To construct a CI for a difference of proportion, we can apply the formulas

$S E = \sqrt{\frac{p_{1} (1 - p_{1})}{n_{1}} + \frac{p_{2} (1 - p_{2})}{n_{2}}}, C I = point estimate \pm z^{*} \times S E$

If necessary, we can use the plug-in principle to estimate the standard error.

$S E \approx \sqrt{\frac{{^p}_{1} (1 - {^p}_{1})}{n_{1}} + \frac{{^p}_{2} (1 - {^p}_{2})}{n_{2}}}$

CI for Difference of Proportions ExamplesLet's obtain a 90% CI for the difference of proportions in the marijuana legalization question. Here is the relevant R code:
Result
Code
## [1] 97 23 84 27
## [1] -0.03775321  0.14090636
n1 <- 120
n2 <- 111
p1 <- 97/n1
p2 <- 84/n2
p_diff <- p1 - p2
(c(n1*p1,n1*(1-p1),n2*p2,n2*(1-p2))) # success-failure conditions
SE <- sqrt(((p1*(1-p1))/n1) + ((p2*(1-p2))/n2))
z_90 <- -qnorm(0.05) # find the appropriate z-value for 90% CI
(CI <- p_diff + z_90 * c(-1,1) * SE) # 90% CI
57

CI for Difference of Proportions ExamplesLet's obtain a 90% CI for the difference of proportions in the marijuana legalization question. Here is the relevant R code:
Result
Code
## [1] 97 23 84 27
## [1] -0.03775321  0.14090636
n1 <- 120
n2 <- 111
p1 <- 97/n1
p2 <- 84/n2
p_diff <- p1 - p2
(c(n1*p1,n1*(1-p1),n2*p2,n2*(1-p2))) # success-failure conditions
SE <- sqrt(((p1*(1-p1))/n1) + ((p2*(1-p2))/n2))
z_90 <- -qnorm(0.05) # find the appropriate z-value for 90% CI
(CI <- p_diff + z_90 * c(-1,1) * SE) # 90% CI
Let's try some more examples together. 
58

Hypotheses for Difference of Proportion TestsThe null hypothesis for a difference of proportions test is typically stated as H0:p1−p2=p0H0:p1−p2=p0 where p0p0 is a hypothetical value for pp. 
59

Hypotheses for Difference of Proportion Tests

The null hypothesis for a difference of proportions test is typically stated as $H_{0} : p_{1} - p_{2} = p_{0}$ where $p_{0}$ is a hypothetical value for $p$ .
The corresponding alternative hypothesis is then one of

Hypotheses for Difference of Proportion Tests

The null hypothesis for a difference of proportions test is typically stated as $H_{0} : p_{1} - p_{2} = p_{0}$ where $p_{0}$ is a hypothetical value for $p$ .
The corresponding alternative hypothesis is then one of
- $H_{A} : p_{1} - p_{2} \neq p_{0}$ (two-sided),

Hypotheses for Difference of Proportion Tests

The null hypothesis for a difference of proportions test is typically stated as $H_{0} : p_{1} - p_{2} = p_{0}$ where $p_{0}$ is a hypothetical value for $p$ .
The corresponding alternative hypothesis is then one of
- $H_{A} : p_{1} - p_{2} \neq p_{0}$ (two-sided),
- $H_{A} : p_{1} - p_{2} < p_{0}$ (one-sided less than), or

Hypotheses for Difference of Proportion Tests

The null hypothesis for a difference of proportions test is typically stated as $H_{0} : p_{1} - p_{2} = p_{0}$ where $p_{0}$ is a hypothetical value for $p$ .
The corresponding alternative hypothesis is then one of
- $H_{A} : p_{1} - p_{2} \neq p_{0}$ (two-sided),
- $H_{A} : p_{1} - p_{2} < p_{0}$ (one-sided less than), or
- $H_{A} : p_{1} - p_{2} > p_{0}$ (one-sided greater than)

Hypotheses for Difference of Proportion Tests

The null hypothesis for a difference of proportions test is typically stated as $H_{0} : p_{1} - p_{2} = p_{0}$ where $p_{0}$ is a hypothetical value for $p$ .
The corresponding alternative hypothesis is then one of
- $H_{A} : p_{1} - p_{2} \neq p_{0}$ (two-sided),
- $H_{A} : p_{1} - p_{2} < p_{0}$ (one-sided less than), or
- $H_{A} : p_{1} - p_{2} > p_{0}$ (one-sided greater than)
When $p_{0} = 0$ we need to use the pooled proportion.

Pooled Proportion

When the null hypothesis is that the proportions are equal, use the pooled proportion ${^p}_{pooled}$ , where

${^p}_{pooled} = \frac{number of "successes"}{number of cases} = \frac{{^p}_{1} n_{1} + {^p}_{2} n_{2}}{n_{1} + n_{2}}$

Pooled Proportion

When the null hypothesis is that the proportions are equal, use the pooled proportion ${^p}_{pooled}$ , where

${^p}_{pooled} = \frac{number of "successes"}{number of cases} = \frac{{^p}_{1} n_{1} + {^p}_{2} n_{2}}{n_{1} + n_{2}}$

The pooled proportion ${^p}_{pooled}$ is used to check the success-failure condition and to estimate the standard error.

Hypothesis Test for Difference of Proportions

Let's apply the hypothesis testing framework to this problem:

An online poll on January 10, 2017 reported that 97 out of 120 people in Virginia between the ages of 18 and 29 believe that marijuana should be legal, while 84 out of 111 who are 30 and over held this belief. Is there a difference between the proportion of young people who favor marijuana leagalization as compared to people who are older?

## [1] 69.73593 19.26407

## [1] 0.9510127

## [1] 0.3415979

n1 <- 120; n2 <- 111
p1 <- 97/120; p2 <- 84/111; p_diff <- p1 - p2
p_pooled <- (p1*n1+p2*n2)/(n1+n2)
(c(n*p_pooled,n*(1-p_pooled)))
SE <- sqrt((p_pooled*(1-p_pooled))/n1 + (p_pooled*(1-p_pooled))/n2)
(z_val <- (p_diff - 0.0)/SE)
(p_value <- 2*(1-pnorm(z_val)))

R Commands for Difference of Proportion TestThe R command prop.test also works for testing a difference of proportions. 
68

R Commands for Difference of Proportion Test

The R command prop.test also works for testing a difference of proportions.
For example, we could solve the marijuana legalization problem with the following:

## 
##     2-sample test for equality of proportions without continuity correction
## 
## data:  c(97, 84) out of c(120, 111)
## X-squared = 0.90443, df = 1, p-value = 0.3416
## alternative hypothesis: two.sided
## 95 percent confidence interval:
##  -0.05486643  0.15801958
## sample estimates:
##    prop 1    prop 2 
## 0.8083333 0.7567568

prop.test(c(97,84),c(120,111),correct = FALSE)

Hypothesis Test for Difference of Proportions ExamplesLet's work some more problems together. 
70

Inference for a Sample MeanA potato chip manufacturer claims that there is on average 32 chips per bag for their brand. How can we tell if this is an accurate claim? 
71

Inference for a Sample Mean

A potato chip manufacturer claims that there is on average 32 chips per bag for their brand. How can we tell if this is an accurate claim?
One approach is to take a sample of, say 25 bags of chips for the particular band and compute the sample mean number of chips per bag.

Inference for a Sample Mean

A potato chip manufacturer claims that there is on average 32 chips per bag for their brand. How can we tell if this is an accurate claim?
One approach is to take a sample of, say 25 bags of chips for the particular band and compute the sample mean number of chips per bag.
This type of problem is inference for a sample mean.

Inference for a Sample Mean

A potato chip manufacturer claims that there is on average 32 chips per bag for their brand. How can we tell if this is an accurate claim?
One approach is to take a sample of, say 25 bags of chips for the particular band and compute the sample mean number of chips per bag.
This type of problem is inference for a sample mean.
In principle, we know the sampling distribution for the sample mean is very nearly normal, provided the conditions for the CLT holds.

Inference for a Sample Mean

A potato chip manufacturer claims that there is on average 32 chips per bag for their brand. How can we tell if this is an accurate claim?
One approach is to take a sample of, say 25 bags of chips for the particular band and compute the sample mean number of chips per bag.
This type of problem is inference for a sample mean.
In principle, we know the sampling distribution for the sample mean is very nearly normal, provided the conditions for the CLT holds.
One question is, what are the appropriate conditions to check to make sure the CLT holds for the sample mean?

Inference for a Sample Mean

A potato chip manufacturer claims that there is on average 32 chips per bag for their brand. How can we tell if this is an accurate claim?
One approach is to take a sample of, say 25 bags of chips for the particular band and compute the sample mean number of chips per bag.
This type of problem is inference for a sample mean.
In principle, we know the sampling distribution for the sample mean is very nearly normal, provided the conditions for the CLT holds.
One question is, what are the appropriate conditions to check to make sure the CLT holds for the sample mean?
When the conditions for the CLT for the sample mean hold, the expression for the standard error involves the population standard deviation ( $σ$ ) which we rarely know if practice. So, how do we estimate the standard error for the sample mean?

Inference for a Sample Mean

A potato chip manufacturer claims that there is on average 32 chips per bag for their brand. How can we tell if this is an accurate claim?
One approach is to take a sample of, say 25 bags of chips for the particular band and compute the sample mean number of chips per bag.
This type of problem is inference for a sample mean.
In principle, we know the sampling distribution for the sample mean is very nearly normal, provided the conditions for the CLT holds.
One question is, what are the appropriate conditions to check to make sure the CLT holds for the sample mean?
When the conditions for the CLT for the sample mean hold, the expression for the standard error involves the population standard deviation ( $σ$ ) which we rarely know if practice. So, how do we estimate the standard error for the sample mean?
We address these questions in the next few slides.

Inference for a Mean VideoYou are encouraged to watch this video on inference for a mean.

78

Conditions for CLT for Sample MeanTwo conditions are required to apply the CLT for a sample mean ˉx¯x:
79

Conditions for CLT for Sample Mean

Two conditions are required to apply the CLT for a sample mean $¯ x$ :

Independence: The sample observations must be independent. Simple random samples from a population are independent, as are data from a random process like rolling a die or tossing a coin.

Conditions for CLT for Sample Mean

Two conditions are required to apply the CLT for a sample mean $¯ x$ :

Independence: The sample observations must be independent. Simple random samples from a population are independent, as are data from a random process like rolling a die or tossing a coin. Normality: When a sample is small, we also require that the sample observations come from a normally distributed population. This condition can be relaxed for larger sample sizes.

Conditions for CLT for Sample Mean

Two conditions are required to apply the CLT for a sample mean $¯ x$ :

Independence: The sample observations must be independent. Simple random samples from a population are independent, as are data from a random process like rolling a die or tossing a coin. Normality: When a sample is small, we also require that the sample observations come from a normally distributed population. This condition can be relaxed for larger sample sizes.

The normality condition is vague but there are some approximate rules that work well in practice.

Practical Check for Normality n<30n<30: If the sample size nn is less than 30 and there are no clear outliers, then we can assume the data come from a nearly normal distribution. 
83

Practical Check for Normality

$n < 30$ : If the sample size $n$ is less than 30 and there are no clear outliers, then we can assume the data come from a nearly normal distribution.
$n \geq 30$ : If the sample size $n$ is at least 30 and there are no particularly exterme outliers, then we can assume the sampling distribution of $¯ x$ is nearly normal, even if the underlying distribution of individual observations is not.

Practical Check for Normality

$n < 30$ : If the sample size $n$ is less than 30 and there are no clear outliers, then we can assume the data come from a nearly normal distribution.
$n \geq 30$ : If the sample size $n$ is at least 30 and there are no particularly exterme outliers, then we can assume the sampling distribution of $¯ x$ is nearly normal, even if the underlying distribution of individual observations is not.
Let's consider example 7.1 from the textbook to illustrate these practical rules.

Estimating Standard ErrorWhen samples are independent and the normality condition is met, then the sampling distribution for the sample mean ˉx¯x is (very nearly) normal with mean μˉx=μμ¯x=μ and standard error SEˉx=σ√nSE¯x=σ√n, where μμ is the true population mean of the distribution from which the samples are taken and σσ is the true population from which the samples are taken. 
86

Estimating Standard Error

When samples are independent and the normality condition is met, then the sampling distribution for the sample mean $¯ x$ is (very nearly) normal with mean $μ_{¯ x} = μ$ and standard error $S E_{¯ x} = \frac{σ}{\sqrt{n}}$ , where $μ$ is the true population mean of the distribution from which the samples are taken and $σ$ is the true population from which the samples are taken.
In practice we do not know the true values of the population parameters $μ$ and $σ$ . But we can use the plug-in principle to obtain estimates:

$μ_{¯ x} \approx ¯ x, and S E_{¯ x} \approx \frac{s}{\sqrt{n}},$

where $s$ is the sample standard deviation.

T-score

If we take $n$ samples from a normal distribution $N (μ, σ)$ , then the quantity

$z = \frac{¯ x - μ}{\frac{σ}{\sqrt{n}}}$ will follow a standard normal distribution $N (0, 1)$ .

T-score

If we take $n$ samples from a normal distribution $N (μ, σ)$ , then the quantity

$z = \frac{¯ x - μ}{\frac{σ}{\sqrt{n}}}$ will follow a standard normal distribution $N (0, 1)$ .

On the other hand, the quantity

$t = \frac{¯ x - μ}{\frac{s}{\sqrt{n}}}$ will not quite be $N (0, 1)$ , especially if $n$ is small.

T-score

If we take $n$ samples from a normal distribution $N (μ, σ)$ , then the quantity

$z = \frac{¯ x - μ}{\frac{σ}{\sqrt{n}}}$ will follow a standard normal distribution $N (0, 1)$ .

On the other hand, the quantity

$t = \frac{¯ x - μ}{\frac{s}{\sqrt{n}}}$ will not quite be $N (0, 1)$ , especially if $n$ is small.

We call the quantity in the last formula a T-score.

T-Score VideoLet's watch this video on T-scores. 

91

t-distributionT-scores follow a t-distribution with degrees of freedom df=n−1df=n−1, where nn is the sample size. 
92

t-distribution

T-scores follow a t-distribution with degrees of freedom $df = n - 1$ , where $n$ is the sample size.
We will use t-distributions for inference, that is, for obtaining confidence intervals and hypothesis testing.

t-distribution

T-scores follow a t-distribution with degrees of freedom $df = n - 1$ , where $n$ is the sample size.
We will use t-distributions for inference, that is, for obtaining confidence intervals and hypothesis testing.
Let's get a feel for t-distributions.

Plotting t-Distributions

We can easily plot density curves for t-distributions.

gf_dist("t",df=19)

Degrees of Freedom Parameter

This plot shows three different t density functions for three different degrees of freedom:

A Computational Experiment

This plot shows a histogram of T-score values obtained by computing the sample mean with $n = 8$ from $N (0, 1)$ . We overlay a density curve for both $N (0, 1)$ and a t-distribution with $df = 7$ :

Area Under a t-Distribution

We can compute areas under t-distribution density curves in the same way we did for areas under normal distribution density curves. For example, the area

is computed by

pt(-1.0,df=5)

## [1] 0.1816087

Middle 95% Under a t DensityHow do we find the middle 95% of area under a t-distribution? This is an important question because it relates to construting a 95% confidence interval for the sample mean. 
99

Middle 95% Under a t Density

How do we find the middle 95% of area under a t-distribution? This is an important question because it relates to construting a 95% confidence interval for the sample mean.
Suppose we have a t-distribution with degrees of freedom $df = 10$ , then to find the value $t^{*}$ so that 95% of the area under the density curve lies between $- t^{*}$ and $t^{*}$ , we use the qt command, for example,

## [1] 2.228139

(t_ast <- -qt(0.05/2,df=10))

100

Middle 95% Under a t Density

How do we find the middle 95% of area under a t-distribution? This is an important question because it relates to construting a 95% confidence interval for the sample mean.
Suppose we have a t-distribution with degrees of freedom $df = 10$ , then to find the value $t^{*}$ so that 95% of the area under the density curve lies between $- t^{*}$ and $t^{*}$ , we use the qt command, for example,

## [1] 2.228139

(t_ast <- -qt(0.05/2,df=10))

We can check our answer:

## [1] 0.95

pt(t_ast,df=10) - pt(-t_ast,df=10)

101

Middle 95% Under a t Density

How do we find the middle 95% of area under a t-distribution? This is an important question because it relates to construting a 95% confidence interval for the sample mean.
Suppose we have a t-distribution with degrees of freedom $df = 10$ , then to find the value $t^{*}$ so that 95% of the area under the density curve lies between $- t^{*}$ and $t^{*}$ , we use the qt command, for example,

## [1] 2.228139

(t_ast <- -qt(0.05/2,df=10))

We can check our answer:

## [1] 0.95

pt(t_ast,df=10) - pt(-t_ast,df=10)

Note that you always have to specify the appropriate value for the degrees of freedom df. The appropriate degrees of freedom is given by $df = n - 1$ , where $n$ is the sample size.

102

One-Sample t CI

Based on a sample of $n$ independent and nearly normal observations, a confidence interval for the population mean is $¯ x \pm t_{df}^{*} \times \frac{s}{\sqrt{n}},$

where is the sample size, is the sample mean, is the sample standard deviation.

103

One-Sample t CI

Based on a sample of independent and nearly normal observations, a confidence interval for the population mean is

where is the sample size, is the sample mean, is the sample standard deviation.

We determine the appropriate value for with the R command

qt((1.0-confidence_level)/2,df=n-1).

104

One-Sample Mean CI Example

Suppose we take a random sample of 13 observations from a normally distributed population and determine the sample mean is 8 with a sample standard deviation of 2.5. Then to find a 90% confidence interval, we would do the following:

## [1] 6.764206 9.235794

n <- 13
x_bar <- 8
s <- 2.5
SE <- s/sqrt(n)
t_ast <- -qt((1.0-0.9)/2,df=n-1)
(CI <- x_bar + t_ast * c(-1,1)*SE)

A 95% CI would be

## [1] 6.489265 9.510735

t_ast <- -qt((1.0-0.95)/2,df=n-1)
(CI <- x_bar + t_ast * c(-1,1)*SE)

105

CI for Single Mean Summary

Once you have determined a one-mean confidence interval would be helpful for an application, there are four steps to constructing the interval:

106

CI for Single Mean Summary

Once you have determined a one-mean confidence interval would be helpful for an application, there are four steps to constructing the interval:

Identify , , and , and determine what confidence level you wish to use.

107

CI for Single Mean Summary

Once you have determined a one-mean confidence interval would be helpful for an application, there are four steps to constructing the interval:

Identify , , and , and determine what confidence level you wish to use.

Verify the conditions to ensure is nearly normal.

108

CI for Single Mean Summary

Once you have determined a one-mean confidence interval would be helpful for an application, there are four steps to constructing the interval:

Identify , , and , and determine what confidence level you wish to use.

Verify the conditions to ensure is nearly normal.
If the conditions hold, approximate by , find , and construct the interval.

109

CI for Single Mean Summary

Once you have determined a one-mean confidence interval would be helpful for an application, there are four steps to constructing the interval:

Identify , , and , and determine what confidence level you wish to use.

Verify the conditions to ensure is nearly normal.
If the conditions hold, approximate by , find , and construct the interval.
Interpret the confidence interval in the context of the problem.

110

One-Mean Hypothesis TestingThe null hypothesis for a one-proportion test is typically stated as H0:μ=μ0 where μ0 is the null value for the population mean. 
111

One-Mean Hypothesis Testing

The null hypothesis for a one-proportion test is typically stated as where is the null value for the population mean.
The corresponding alternative hypothesis is then one of

112

One-Mean Hypothesis Testing

The null hypothesis for a one-proportion test is typically stated as where is the null value for the population mean.
The corresponding alternative hypothesis is then one of
- (two-sided),

113

One-Mean Hypothesis Testing

The null hypothesis for a one-proportion test is typically stated as where is the null value for the population mean.
The corresponding alternative hypothesis is then one of
- (two-sided),
- (one-sided less than), or

114

One-Mean Hypothesis Testing

The null hypothesis for a one-proportion test is typically stated as where is the null value for the population mean.
The corresponding alternative hypothesis is then one of
- (two-sided),
- (one-sided less than), or
- (one-sided greater than)

115

One-Mean Hypothesis Test ProcedureOnce you have determined a one-mean hypothesis test is the correct procedure, there are four steps to completing the test:
116

One-Mean Hypothesis Test Procedure

Once you have determined a one-mean hypothesis test is the correct procedure, there are four steps to completing the test:
- Identify the parameter of interest, list out hypotheses, identify the significance level, and identify , , and .

117

One-Mean Hypothesis Test Procedure

Once you have determined a one-mean hypothesis test is the correct procedure, there are four steps to completing the test:
- Identify the parameter of interest, list out hypotheses, identify the significance level, and identify , , and .
- Verify conditions to ensure is nearly normal.

118

One-Mean Hypothesis Test Procedure

Once you have determined a one-mean hypothesis test is the correct procedure, there are four steps to completing the test:
- Identify the parameter of interest, list out hypotheses, identify the significance level, and identify , , and .
- Verify conditions to ensure is nearly normal.
- If the conditions hold, approximate by , compute the T-score using
and compute the p-value.

119

One-Mean Hypothesis Test Procedure

Once you have determined a one-mean hypothesis test is the correct procedure, there are four steps to completing the test:
- Identify the parameter of interest, list out hypotheses, identify the significance level, and identify , , and .
- Verify conditions to ensure is nearly normal.
- If the conditions hold, approximate by , compute the T-score using
and compute the p-value.
- Evaluate the hypothesis test by comparing the p-value to the significance level , and provide a conclusion in the context of the problem.

120

A Simple Example

We want to know if the following data comes from a normal distribution with mean 0:

##  [1] -0.36047565 -0.03017749  1.75870831  0.27050839  0.32928774  1.91506499
##  [7]  0.66091621 -1.06506123 -0.48685285 -0.24566197

We can apply a hypothesis test corresponding to , versus as follows

## [1] 0.9105232

## [1] 0.3862831

n <- length(x); x_bar <- mean(x); s <- sd(x)
alpha <- 0.05
SE <- s/sqrt(n)
mu_0 <- 0.0
(T <- (x_bar - mu_0)/SE)
(p_value <- 2*(1-pt(T,df=n-1)))

121

R Command for One-Sample t-testThere is a built-in R command, t.test that will conduct a hypothesis test for a sinlg emean for us. 
122

R Command for One-Sample t-test

There is a built-in R command, t.test that will conduct a hypothesis test for a sinlg emean for us.
For example, we can solve our previous problem using

## 
##     One Sample t-test
## 
## data:  x
## t = 0.91052, df = 9, p-value = 0.3863
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  -0.4076704  0.9569217
## sample estimates:
## mean of x 
## 0.2746256

t.test(x,mu=0.0)

123

R Command for One-Sample t-test

There is a built-in R command, t.test that will conduct a hypothesis test for a sinlg emean for us.
For example, we can solve our previous problem using

## 
##     One Sample t-test
## 
## data:  x
## t = 0.91052, df = 9, p-value = 0.3863
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  -0.4076704  0.9569217
## sample estimates:
## mean of x 
## 0.2746256

t.test(x,mu=0.0)

Let's work some examples together.

124

Paired Data

Two sets of observations are paired if each observation in one set has a special correspondence or connection with exactly one observation in the other data set.

125

Paired Data

Two sets of observations are paired if each observation in one set has a special correspondence or connection with exactly one observation in the other data set.

Common examples of paired data correspond to "before" and "after" trials.

126

Paired Data

Two sets of observations are paired if each observation in one set has a special correspondence or connection with exactly one observation in the other data set.

Common examples of paired data correspond to "before" and "after" trials.

For example, does a particular study technique work well for increasing one's exam score? To test this, we can ask 25 people to take an exam and record their scores, then we can ask those same people to try the study technique before taking another similar exam.

127

Paired Data

Two sets of observations are paired if each observation in one set has a special correspondence or connection with exactly one observation in the other data set.

Common examples of paired data correspond to "before" and "after" trials.

For example, does a particular study technique work well for increasing one's exam score? To test this, we can ask 25 people to take an exam and record their scores, then we can ask those same people to try the study technique before taking another similar exam.

As another example, suppose it is claimed that among the general population of adults in the US, the average length of the left foot is longer than the average length of the right foot. To test this, we can select 32 people, record the measurement of the left foot of everyone in one column, then record the measurement of the right foot of everyone in a second column. We must make sure that each row of the resulting data frame corresponds with only one person.

128

Paired Data Example

Perhaps the data from our last example looks as follows:

## # A tibble: 6 × 3
##   left_foot right_foot foot_diff
##       <dbl>      <dbl>     <dbl>
## 1      7.83       8.27    -0.437
## 2      7.93       8.26    -0.332
## 3      8.47       8.25     0.221
## 4      8.02       8.21    -0.185
## 5      8.04       8.17    -0.127
## 6      8.51       7.98     0.533

129

Paired Data Example

Perhaps the data from our last example looks as follows:

## # A tibble: 6 × 3
##   left_foot right_foot foot_diff
##       <dbl>      <dbl>     <dbl>
## 1      7.83       8.27    -0.437
## 2      7.93       8.26    -0.332
## 3      8.47       8.25     0.221
## 4      8.02       8.21    -0.185
## 5      8.04       8.17    -0.127
## 6      8.51       7.98     0.533

Notice that we have added a column that is the difference of the left foot measurement minus the right foot measurement.

130

Hypothesis Test for Paired DataHow can we set up a hypothesis testing framework for the foot measurement question? 
131

Hypothesis Test for Paired Data

How can we set up a hypothesis testing framework for the foot measurement question?
Basically, we can apply statistical inference to the difference column in the data.

132

Hypothesis Test for Paired Data

How can we set up a hypothesis testing framework for the foot measurement question?
Basically, we can apply statistical inference to the difference column in the data.
The typical null hypothesis for paired data is that the average difference between the measurements is 0. We write this as

133

Paired Hypothesis Test ProcedureOnce you have determined a paired hypothesis test is the correct procedure, there are four steps to completing the test:
134

Paired Hypothesis Test Procedure

Once you have determined a paired hypothesis test is the correct procedure, there are four steps to completing the test:
- Determine the significance level, the sample size , the mean of the differences , and the corresponding standard deviation .

135

Paired Hypothesis Test Procedure

Once you have determined a paired hypothesis test is the correct procedure, there are four steps to completing the test:
- Determine the significance level, the sample size , the mean of the differences , and the corresponding standard deviation .
- Verify the conditions to ensure that is nearly normal.

136

Paired Hypothesis Test Procedure

Once you have determined a paired hypothesis test is the correct procedure, there are four steps to completing the test:
- Determine the significance level, the sample size , the mean of the differences , and the corresponding standard deviation .
- Verify the conditions to ensure that is nearly normal.
- If the conditions hold, approximate by , compute the T-score using
and compute the p-value.

137

Paired Hypothesis Test Procedure

Once you have determined a paired hypothesis test is the correct procedure, there are four steps to completing the test:
- Determine the significance level, the sample size , the mean of the differences , and the corresponding standard deviation .
- Verify the conditions to ensure that is nearly normal.
- If the conditions hold, approximate by , compute the T-score using
and compute the p-value.
- Evaluate the hypothesis test by comparing the p-value to the significance level , and provide a conclusion in the context of the problem.

138

First ExampleConsider our foot measurement data.The sample size is n=32 and a boxplot shows that there are no extreme outliers. Now we compute the necessary quantities:
Result
Code
## [1] -0.6907872
## [1] 0.4948396
n <- 32
d_bar <- mean(foot_df$foot_diff)
s_d <- sd(foot_df$foot_diff)
(t_val <- d_bar/(s_d/sqrt(n)))
(p_val <- 2*pt(t_val,df=n-1))
139

First ExampleConsider our foot measurement data.The sample size is n=32 and a boxplot shows that there are no extreme outliers. Now we compute the necessary quantities:
Result
Code
## [1] -0.6907872
## [1] 0.4948396
n <- 32
d_bar <- mean(foot_df$foot_diff)
s_d <- sd(foot_df$foot_diff)
(t_val <- d_bar/(s_d/sqrt(n)))
(p_val <- 2*pt(t_val,df=n-1))
Here we fail to reject the null hypothesis at the 0.05 significance level. That is, the data does not provide sufficient evidence for rejecting the null hypothesis that there is no significant difference in the length of the left foot versus the right foot.
140

R Command for Paired Hypothesis TestAgain, we can use the t.test function. However, now we must use two sets of data and add the paired=TRUE argument. 
141

R Command for Paired Hypothesis Test

Again, we can use the t.test function. However, now we must use two sets of data and add the paired=TRUE argument.
For example, to test the hypothesis for the foot data, one would use

## 
##     Paired t-test
## 
## data:  foot_df$left_foot and foot_df$right_foot
## t = -0.69079, df = 31, p-value = 0.4948
## alternative hypothesis: true mean difference is not equal to 0
## 95 percent confidence interval:
##  -0.18623480  0.09199711
## sample estimates:
## mean difference 
##     -0.04711885

t.test(foot_df$left_foot,foot_df$right_foot,paired = TRUE)

142

R Command for Paired Hypothesis Test

Again, we can use the t.test function. However, now we must use two sets of data and add the paired=TRUE argument.
For example, to test the hypothesis for the foot data, one would use

## 
##     Paired t-test
## 
## data:  foot_df$left_foot and foot_df$right_foot
## t = -0.69079, df = 31, p-value = 0.4948
## alternative hypothesis: true mean difference is not equal to 0
## 95 percent confidence interval:
##  -0.18623480  0.09199711
## sample estimates:
## mean difference 
##     -0.04711885

t.test(foot_df$left_foot,foot_df$right_foot,paired = TRUE)

Let's work some more examples together.

143

Difference of Two MeansInference for paired data compares two different (but related) measurements on the same population. For example, we might want to study the difference in how individuals sleep before and after consuming a large amount of caffeine. 
144

Difference of Two Means

Inference for paired data compares two different (but related) measurements on the same population. For example, we might want to study the difference in how individuals sleep before and after consuming a large amount of caffeine.
On the other hand, inference for a difference of two means compares the same measurement on two difference populations. For example, we may want to study any difference between how caffeine affects the sleep of individuals with high blood pressure compared to those that do not have high blood pressure.

145

Difference of Two Means

Inference for paired data compares two different (but related) measurements on the same population. For example, we might want to study the difference in how individuals sleep before and after consuming a large amount of caffeine.
On the other hand, inference for a difference of two means compares the same measurement on two difference populations. For example, we may want to study any difference between how caffeine affects the sleep of individuals with high blood pressure compared to those that do not have high blood pressure.
We can still use a t-distribution for inference for a difference of two means but we must compute the two-sample means and standard deviations separately for estimating standard error.

146

Difference of Two Means

Inference for paired data compares two different (but related) measurements on the same population. For example, we might want to study the difference in how individuals sleep before and after consuming a large amount of caffeine.
On the other hand, inference for a difference of two means compares the same measurement on two difference populations. For example, we may want to study any difference between how caffeine affects the sleep of individuals with high blood pressure compared to those that do not have high blood pressure.
We can still use a t-distribution for inference for a difference of two means but we must compute the two-sample means and standard deviations separately for estimating standard error.

We proceed with the details.

147

Confidence Intervals for a Difference in MeansThe t-distribution can be used for inference when working with the standardized difference of two means if
148

Confidence Intervals for a Difference in Means

The t-distribution can be used for inference when working with the standardized difference of two means if
- The data are independent within and between the two groups, e.g., the data come from independent random samples or from a randomized experiment.

149

Confidence Intervals for a Difference in Means

The t-distribution can be used for inference when working with the standardized difference of two means if
- The data are independent within and between the two groups, e.g., the data come from independent random samples or from a randomized experiment.
- We check the outliers rules of thumb for each group separately.

150

Confidence Intervals for a Difference in Means

The t-distribution can be used for inference when working with the standardized difference of two means if
- The data are independent within and between the two groups, e.g., the data come from independent random samples or from a randomized experiment.
- We check the outliers rules of thumb for each group separately.
The standard error may be computed as

151

Hypothesis Tests for Difference of MeansHypothesis tests for a difference of two means works in a very similar fashion to what we have seen before. 
152

Hypothesis Tests for Difference of Means

Hypothesis tests for a difference of two means works in a very similar fashion to what we have seen before.
To conduct a test for a difference of two means "by hand", use the smaller of the two degrees of freedom.

153

Hypothesis Tests for Difference of Means

Hypothesis tests for a difference of two means works in a very similar fashion to what we have seen before.
To conduct a test for a difference of two means "by hand", use the smaller of the two degrees of freedom.
Use the t.test function without the paired = TRUE argument.

154

Examples of Inference for Difference of MeansLet's look at some examples and work some problems together. 
155

Statistical PowerOur next topic is statistical power, see the following video to get an introduction. 

156

Notes157

Notes158

Notes159

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