# 10.1. Hypothesis Testing: Two Population Means and Two Population Proportions*

## Student Learning Objectives

By the end of this chapter, the student should be able to:

• Classify hypothesis tests by type.

• Conduct and interpret hypothesis tests for two population means, population standard deviations known.

• Conduct and interpret hypothesis tests for two population means, population standard deviations unknown.

• Conduct and interpret hypothesis tests for two population proportions.

• Conduct and interpret hypothesis tests for matched or paired samples.

## Introduction

Studies often compare two groups. For example, researchers are interested in the effect aspirin has in preventing heart attacks. Over the last few years, newspapers and magazines have reported about various aspirin studies involving two groups. Typically, one group is given aspirin and the other group is given a placebo. Then, the heart attack rate is studied over several years.

There are other situations that deal with the comparison of two groups. For example, studies compare various diet and exercise programs. Politicians compare the proportion of individuals from different income brackets who might vote for them. Students are interested in whether SAT or GRE preparatory courses really help raise their scores.

In the previous chapter, you learned to conduct hypothesis tests on single means and single proportions. You will expand upon that in this chapter. You will compare two averages or two proportions to each other. The general procedure is still the same, just expanded.

To compare two averages or two proportions, you work with two groups. The groups are classified either as independent or matched pairs. Independent groups mean that the two samples taken are independent, that is, sample values selected from one population are not related in any way to sample values selected from the other population. Matched pairs consist of two samples that are dependent. The parameter tested using matched pairs is the population mean. The parameters tested using independent groups are either population means or population proportions.

### Note

This chapter relies on either a calculator or a computer to calculate the degrees of freedom, the test statistics, and p-values. TI-83+ and TI-84 instructions are included as well as the test statistic formulas. Because of technology, we do not need to separate two population means, independent groups, population variances unknown into large and small sample sizes.

This chapter deals with the following hypothesis tests:

Independent groups (samples are independent)

• Test of two population means.

• Test of two population proportions.

Matched or paired samples (samples are dependent)

• Becomes a test of one population mean.

# 10.2. Comparing Two Independent Population Means with Unknown Population Standard Deviations*

1. The two independent samples are simple random samples from two distinct populations.

2. Both populations are normally distributed with the population means and standard deviations unknown unless the sample sizes are greater than 30. In that case, the populations need not be normally distributed.

The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual samples. In order to account for the variation, we take the difference of the sample means, - , and divide by the standard error (shown below) in order to standardize the difference. The result is a t-score test statistic (shown below).

Because we do not know the population standard deviations, we estimate them using the two sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error, of the difference in sample means, - .

(10.1)
The standard error is: The test statistic (t-score) is calculated as follows:

(10.2)
T-score where:

• s 1 and s 2 , the sample standard deviations, are estimates of σ 1 and σ 2 , respectively.

• σ 1 and σ 2 are the unknown population standard deviations.

• and are the sample means. μ 1 and μ 2 are the population means.

The degrees of freedom (df) is a somewhat complicated calculation. However, a computer or calculator calculates it easily. The dfs are not always a whole number. The test statistic calculated above is approximated by the Student-t distribution with dfs as follows:

(10.3)
Degrees of freedom When both sample sizes n 1 and n 2 are five or larger, the Student-t approximation is very good. Notice that the sample variances s 1 2 and s 2 2 are not pooled. (If the question comes up, do not pool the variances.)

### Note

It is not necessary to compute this by hand. A calculator or computer easily computes it.

Example 10.1. Independent groups

The average amount of time boys and girls ages 7 through 11 spend playing sports each day is believed to be the same. An experiment is done, data is collected, resulting in the table below:

Table 10.1.
Sample SizeAverage Number of Hours Playing Sports Per DaySample Standard Deviation
Girls92 hours Boys163.2 hours1.00

Problem

Is there a difference in the average amount of time boys and girls ages 7 through 11 play sports each day? Test at the 5% level of significance.

Solution

The population standard deviations are not known. Let g be the subscript for girls and b be the subscript for boys. Then, μ g is the population mean for girls and μ b is the population mean for boys. This is a test of two independent groups, two population means.

Random variable: = difference in the average amount of time girls and boys play sports each day.

H o : H a : The words “the same” tell you H o has an “=”. Since there are no other words to indicate H a , then assume “is different.” This is a two-tailed test.

Distribution for the test: Use t df where df is calculated using the df formula for independent groups, two population means. Using a calculator, df is approximately 18.8462. Do not pool the variances.

Calculate the p-value using a Student-t distribution: p-value = 0.0054

Graph:

Figure 10.1.  s b = 1

So, Half the p-value is below -1.2 and half is above 1.2.

Make a decision: Since α > p-value, reject H o .

This means you reject μ g = μ b . The means are different.

Conclusion: At the 5% level of significance, the sample data show there is sufficient evidence to conclude that the average number of hours that girls and boys aged 7 through 11 play sports per day is different.

### Note

TI-83+ and TI-84: Press `STAT`. Arrow over to `TESTS` and press `4:2-SampTTest`. Arrow over to Stats and press `ENTER`. Arrow down and enter `2` for the first sample mean, ``` ``` for Sx1, `9` for n1, `3.2` for the second sample mean, `1` for Sx2, and `16` for n2. Arrow down to μ1: and arrow to `does not equal` μ2. Press `ENTER`. Arrow down to Pooled: and `No`. Press `ENTER`. Arrow down to `Calculate` and press `ENTER`. The p-value is p = 0.0054, the dfs are approximately 18.8462, and the test statistic is -3.14. Do the procedure again but instead of Calculate do Draw.

Example 10.2.

A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is 4 math classes with a standard deviation of 1.5 math classes. College B samples 9 graduates. Their average is 3.5 math classes with a standard deviation of 1 math class. The community group believes that a student who graduates from college A has taken more math classes, on the average. Test at a 1% significance level. Answer the following questions.

Problem 1. (Go to Solution)

Is this a test of two means or two proportions?

Problem 2. (Go to Solution)

Are the populations standard deviations known or unknown?

Problem 3. (Go to Solution)

Which distribution do you use to perform the test?

Problem 4. (Go to Solution)

What is the random variable?

Problem 5. (Go to Solution)

What are the null and alternate hypothesis?

Problem 6. (Go to Solution)

Is this test right, left, or two tailed?

Problem 7. (Go to Solution)

What is the p-value?

Problem 8. (Go to Solution)

Do you reject or not reject the null hypothesis?

Conclusion:

At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a student who graduates from college A has taken more math classes, on the average, than a student who graduates from college B.

## Solutions to Exercises

two means

unknown

Student-t • H o : μ A μ B

• H a : μ A > μ B

right

0.1928

Do not reject.

## Glossary

Degrees of Freedom (df)

The number of objects in a sample that are free to vary.

Standard Deviation

A number that is equal to the square root of the variance and measures how far data values are from their mean. Notation: s for sample standard deviation and σ for population standard deviation.

Variable (Random Variable)

A characteristic of interest in a population being studied. Common notation for variables are upper case Latin letters X , Y , Z ,…; common notation for a specific value from the domain (set of all possible values of a variable) are lower case Latin letters x , y , z ,…. For example, if X is the number of children in a family, then x represents a specific integer 0, 1, 2, 3, …. Variables in statistics differ from variables in intermediate algebra in two following ways.

• The domain of the random variable (RV) is not necessarily a numerical set; the domain may be expressed in words; for example, if X = hair color then the domain is {black, blond, gray, green, orange}.

• We can tell what specific value x of the Random Variable X takes only after performing the experiment.

# 10.3. Comparing Two Independent Population Means with Known Population Standard Deviations*

Even though this situation is not likely (knowing the population standard deviations is not likely), the following example illustrates hypothesis testing for independent means, known population standard deviations. The distribution is Normal and is for the difference of sample means, . The normal distribution has the following format:

(10.4)
Normal distribution (10.5)
The standard deviation is: (10.6)
The test statistic (z-score) is: Example 10.3.

independent groups, population standard deviations known: The mean lasting time of 2 competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax. The following table is the result.

Table 10.2.
WaxSample Mean Number of Months Floor Wax LastPopulation Standard Deviation
130.33
22.90.36

Problem

Does the data indicate that wax 1 is more effective than wax 2? Test at a 5% level of significance.

Solution

This is a test of two independent groups, two population means, population standard deviations known.

Random Variable: difference in the average number of months the competing floor waxes last.

H o : μ 1μ 2

H a : μ 1 > μ 2

The words “is more effective” says that wax 1 lasts longer than wax 2, on the average. “Longer” is a ” > “ symbol and goes into H a . Therefore, this is a right-tailed test.

Distribution for the test: The population standard deviations are known so the distribution is normal. Using the formula above, the distribution is: Since μ 1 μ 2 then μ 1 μ 2 ≤ 0 and the mean for the normal distribution is 0.

Calculate the p-value using the normal distribution: p-value = 0.1799

Graph:

Figure 10.2.  Compare α and the p-value: α = 0.05 and p-value = 0.1799. Therefore, α <  p-value.

Make a decision: Since α < p-value, do not reject H o .

Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that wax 1 lasts longer (wax 1 is more effective) than wax 2.

### Note

TI-83+ and TI-84: Press `STAT`. Arrow over to `TESTS` and press `3:2-SampZTest`. Arrow over to `Stats` and press `ENTER`. Arrow down and enter `.33` for sigma1, `.36` for sigma2, `3` for the first sample mean, `20` for n1, `2.9` for the second sample mean, and `20` for n2. Arrow down to μ 1: and arrow to > μ 2. Press `ENTER`. Arrow down to `Calculate` and press `ENTER`. The p-value is p = 0.1799 and the test statistic is 0.9157. Do the procedure again but instead of `Calculate` do `Draw`.

# 10.4. Comparing Two Independent Population Proportions*

1. The two independent samples are simple random samples that are independent.

2. The number of successes is at least five and the number of failures is at least five for each of the samples.

Comparing two proportions, like comparing two means, is common. If two estimated proportions are different, it may be due to a difference in the populations or it may be due to chance. A hypothesis test can help determine if a difference in the estimated proportions reflects a difference in the populations.

The difference of two proportions follows an approximate normal distribution. Generally, the null hypothesis states that the two proportions are the same. That is, H o : p A = p B . To conduct the test, we use a pooled proportion, p c .

(10.7)
The pooled proportion is calculated as follows: (10.8)
The distribution for the differences is: (10.9)
The test statistic (z-score) is: Example 10.4. Two population proportions

Two types of medication for hives are being tested to determine if there is a difference in the percentage of adult patient reactions. Twenty out of a random sample of 200 adults given medication A still had hives 30 minutes after taking the medication. Twelve out of another random sample of 200 adults given medication B still had hives 30 minutes after taking the medication. Test at a 1% level of significance.

### Determining the solution

This is a test of 2 population proportions.

Problem (Go to Solution)

How do you know?

Let A and B be the subscripts for medication A and medication B. Then p A and p B are the desired population proportions.

Random Variable:

P’ A − P’ B = difference in the percentages of adult patients who did not react after 30 minutes to medication A and medication B.  The words “is a difference” tell you the test is two-tailed.

Distribution for the test: Since this is a test of two binomial population proportions, the distribution is normal: Therefore, P’ A − P’ B follows an approximate normal distribution.

Calculate the p-value using the normal distribution: p-value = 0.1404.

Estimated proportion for group A: Estimated proportion for group B: Graph:

Figure 10.3. P’ A − P’ B = 0.1 – 0.06 = 0.04 .

Half the p-value is below -0.04 and half is above 0.04.

Compare α and the p-value: α = 0.01 and the p-value = 0.1404. α <  p-value.

Make a decision: Since α < p-value , you cannot reject H o .

Conclusion: At a 1% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the percentages of adult patients who did not react after 30 minutes to medication A and medication B.

### Note

TI-83+ and TI-84: Press `STAT`. Arrow over to `TESTS` and press `6:2-PropZTest`. Arrow down and enter `20` for x1, `200` for n1, `12` for x2, and `200` for n2. Arrow down to `p1`: and arrow to `not equal p2`. Press `ENTER`. Arrow down to `Calculate` and press `ENTER`. The p-value is p = 0.1404 and the test statistic is 1.47. Do the procedure again but instead of `Calculate` do `Draw`.

## Solutions to Exercises

The problem asks for a difference in percentages.

# 10.5. Matched or Paired Samples*

1. Simple random sampling is used.

2. Sample sizes are often small.

3. Two measurements (samples) are drawn from the same pair of individuals or objects.

4. Differences are calculated from the matched or paired samples.

5. The differences form the sample that is used for the hypothesis test.

6. The matched pairs have differences that either come from a population that is normal or the number of differences is greater than 30 or both.

In a hypothesis test for matched or paired samples, subjects are matched in pairs and differences are calculated. The differences are the data. The population mean for the differences, μ d , is then tested using a Student-t test for a single population mean with n – 1 degrees of freedom where n is the number of differences.

(10.10)
The test statistic (t-score) is: Example 10.5. Matched or paired samples

A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are shown in the table. The “before” value is matched to an “after” value.

Table 10.3.
Subject:ABCDEFGH
Before6.66.59.010.311.38.16.311.6
After6.82.47.48.58.16.13.42.0

Problem

Are the sensory measurements, on average, lower after hypnotism? Test at a 5% significance level.

Solution

Corresponding “before” and “after” values form matched pairs.

Table 10.4.
6.86.60.2
2.46.5-4.1
7.49-1.6
8.510.3-1.8
8.111.3-3.2
6.18.1-2
3.46.3-2.9
211.6-9.6

The data for the test are the differences: {0.2, -4.1, -1.6, -1.8, -3.2, -2, -2.9, -9.6}

The sample mean and sample standard deviation of the differences are: and s d = 2.91 Verify these values.

Let μ d be the population mean for the differences. We use the subscript d to denote “differences.”

Random Variable: = the average difference of the sensory measurements

H o : μ d ≥ 0 There is no improvement. ( μ d is the population mean of the differences.)

H a : μ d < 0 There is improvement. The score should be lower after hypnotism so the difference ought to be negative to indicate improvement.

Distribution for the test: The distribution is a student-t with df = n – 1 = 8 – 1 = 7 . Use t 7 . (Notice that the test is for a single population mean.)

Calculate the p-value using the Student-t distribution: p-value = 0.0095

Graph:

Figure 10.4.  is the random variable for the differences.

The sample mean and sample standard deviation of the differences are:  Compare α and the p-value: α = 0.05 and p-value = 0.0095 . α > p-value.

Make a decision: Since α > p-value, reject H o .

This means that μ d < 0 and there is improvement.

Conclusion: At a 5% level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower after hypnotism. Hypnotism appears to be effective in reducing pain.

### Note

For the TI-83+ and TI-84 calculators, you can either calculate the differences ahead of time (after - before) and put the differences into a list or you can put the after data into a first list and the before data into a second list. Then go to a third list and arrow up to the name. Enter 1st list name - 2nd list name. The calculator will do the subtraction and you will have the differences in the third list.

### Note

TI-83+ and TI-84: Use your list of differences as the data. Press `STAT` and arrow over to `TESTS`. Press `2:T-Test`. Arrow over to `Data` and press `ENTER`. Arrow down and enter `0` for μ 0 , the name of the list where you put the data, and `1` for Freq:. Arrow down to `μ`: and arrow over to `<` μ 0 . Press `ENTER`. Arrow down to `Calculate` and press `ENTER`. The p-value is 0.0094 and the test statistic is -3.04. Do these instructions again except arrow to `Draw` (instead of `Calculate`). Press `ENTER`.

Example 10.6.

A college football coach was interested in whether the college’s strength development class increased his players’ maximum lift (in pounds) on the bench press exercise. He asked 4 of his players to participate in a study. The amount of weight they could each lift was recorded before they took the strength development class. After completing the class, the amount of weight they could each lift was again measured. The data are as follows:

Table 10.5.
Weight (in pounds)Player 1Player 2Player 3Player 4
Amount of weighted lifted prior to the class205241338368
Amount of weight lifted after the class295252330360

The coach wants to know if the strength development class makes his players stronger, on average.

Problem (Go to Solution)

Record the differences data. Calculate the differences by subtracting the amount of weight lifted prior to the class from the weight lifted after completing the class. The data for the differences are: {90, 11, -8, -8}

Using the differences data, calculate the sample mean and the sample standard deviation. Using the difference data, this becomes a test of a single __________ (fill in the blank).

Define the random variable: average difference in the maximum lift per player.

The distribution for the hypothesis test is t 3 . Graph:

Figure 10.5. Calculate the p-value: The p-value is 0.2150

Decision: If the level of significance is 5%, the decision is to not reject the null hypothesis because α < p-value .

What is the conclusion?

Example 10.7.

Seven eighth graders at Kennedy Middle School measured how far they could push the shot-put with their dominant (writing) hand and their weaker (non-writing) hand. They thought that they could push equal distances with either hand. The following data was collected.

Table 10.6.
Distance (in feet) usingStudent 1Student 2Student 3Student 4Student 5Student 6Student 7
Dominant Hand30263417192620
Weaker Hand28142718172616

Problem (Go to Solution)

Conduct a hypothesis test to determine whether the differences in distances between the children’s dominant versus weaker hands is significant.

### Hint

use a t-test on the difference data.

### Check

The test statistic is 2.18 and the p-value is 0.0716.

What is your conclusion?

## Solutions to Exercises

means; At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the strength development class helped to make the players stronger, on average.

H 0 : μ d equals 0; H a : μ d does not equal 0; Do not reject the null; At a 5% significance level, from the sample data, there is not sufficient evidence to conclude that the differences in distances between the children’s dominant versus weaker hands is significant (there is not sufficient evidence to show that the children could push the shot-put further with their dominant hand). Alpha and the p-value are close so the test is not strong.

# 10.6. Summary of Types of Hypothesis Tests*

Two Population Means

• Populations are independent and population standard deviations are unknown.

• Populations are independent and population standard deviations are known (not likely).

Matched or Paired Samples

• Two samples are drawn from the same set of objects.

• Samples are dependent.

Two Population Proportions

• Populations are independent.

# 10.7. Practice 1: Hypothesis Testing for Two Proportions*

## Student Learning Outcomes

• The student will explore the properties of hypothesis testing with two proportions.

## Given

In the 2000 Census, 2.4 percent of the U.S. population reported being two or more races. However, the percent varies tremendously from state to state. (http://www.census.gov/prod/2001pubs/c2kbr01-6.pdf) Suppose that two random surveys are conducted. In the first random survey, out of 1000 North Dakotans, only 9 people reported being of two or more races. In the second random survey, out of 500 Nevadans, 17 people reported being of two or more races. Conduct a hypothesis test to determine if the population percents are the same for the two states or if the percent for Nevada is statistically higher than for North Dakota.

## Hypothesis Testing: Two Proportions

Exercise 10.7.1. (Go to Solution)

Is this a test of averages or proportions?

Exercise 10.7.2. (Go to Solution)

State the null and alternative hypotheses.

 a. H 0 : b. H a :

Exercise 10.7.3. (Go to Solution)

Is this a right-tailed, left-tailed, or two-tailed test? How do you know?

Exercise 10.7.4.

What is the Random Variable of interest for this test?

Exercise 10.7.5.

In words, define the Random Variable for this test.

Exercise 10.7.6. (Go to Solution)

Which distribution (Normal or student-t) would you use for this hypothesis test?

Exercise 10.7.7.

Explain why you chose the distribution you did for the above question.

Exercise 10.7.8. (Go to Solution)

Calculate the test statistic.

Exercise 10.7.9.

Sketch a graph of the situation. Label the horizontal axis. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the pvalue.

Figure 10.6. Exercise 10.7.10. (Go to Solution)

Find the pvalue:

Exercise 10.7.11. (Go to Solution)

At a pre-conceived α = 0.05, what is your:

 a. Decision: b. Reason for the decision: c. Conclusion (write out in a complete sentence):

## Discussion Question

Exercise 10.7.12.

Does it appear that the proportion of Nevadans who are two or more races is higher than the proportion of North Dakotans? Why or why not?

## Solutions to Exercises

Proportions

 a. a. right-tailed

Normal

3.50

0.0002

 a. Reject the null hypothesis

# 10.8. Practice 2: Hypothesis Testing for Two Averages*

## Student Learning Outcome

• The student will explore the properties of hypothesis testing with two averages.

## Given

The U.S. Center for Disease Control reports that the average life expectancy for whites born in 1900 was 47.6 years and for nonwhites it was 33.0 years. (http://www.cdc.gov/nchs/data/dvs/nvsr53_06t12.pdf ) Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the average life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the average life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the average life spans in the county were the same for whites and nonwhites.

## Hypothesis Testing: Two Averages

Exercise 10.8.1. (Go to Solution)

Is this a test of averages or proportions?

Exercise 10.8.2. (Go to Solution)

State the null and alternative hypotheses.

 a. H 0 : b. H a :

Exercise 10.8.3. (Go to Solution)

Is this a right-tailed, left-tailed, or two-tailed test? How do you know?

Exercise 10.8.4. (Go to Solution)

What is the Random Variable of interest for this test?

Exercise 10.8.5. (Go to Solution)

In words, define the Random Variable of interest for this test.

Exercise 10.8.6.

Which distribution (Normal or student-t) would you use for this hypothesis test?

Exercise 10.8.7.

Explain why you chose the distribution you did for the above question.

Exercise 10.8.8. (Go to Solution)

Calculate the test statistic.

Exercise 10.8.9.

Sketch a graph of the situation. Label the horizontal axis. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the pvalue.

Figure 10.7. Exercise 10.8.10. (Go to Solution)

Find the pvalue:

Exercise 10.8.11. (Go to Solution)

At a pre-conceived α = 0.05, what is your:

 a. Decision: b. Reason for the decision: c. Conclusion (write out in a complete sentence):

## Discussion Question

Exercise 10.8.12.

Does it appear that the averages are the same? Why or why not?

## Solutions to Exercises

Averages

 a. b. two-tailed The difference between the average life spans of whites and nonwhites.

5.42

0.0000

 a. Reject the null hypothesis

# 10.9. Homework*

For questions Exercise 10.9.1. - Exercise 10.9.10., indicate which of the following choices best identifies the hypothesis test.

 A. Independent group means, population standard deviations and/or variances known B. Independent group means, population standard deviations and/or variances unknown C. Matched or paired samples D. Single mean E. 2 proportions F. Single proportion

Exercise 10.9.1. (Go to Solution)

A powder diet is tested on 49 people and a liquid diet is tested on 36 different people. The population standard deviations are 2 pounds and 3 pounds, respectively. Of interest is whether the liquid diet yields a higher average weight loss than the powder diet.

Exercise 10.9.2.

A new chocolate bar is taste-tested on consumers. Of interest is whether the percentage of children that like the new chocolate bar is greater than the percentage of adults that like it.

Exercise 10.9.3. (Go to Solution)

The average number of English courses taken in a two–year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 9 males and 16 females.

Exercise 10.9.4.

A football league reported that the average number of touchdowns per game was 5. A study is done to determine if the average number of touchdowns has decreased.

Exercise 10.9.5. (Go to Solution)

A study is done to determine if students in the California state university system take longer to graduate than students enrolled in private universities. 100 students from both the California state university system and private universities are surveyed. From years of research, it is known that the population standard deviations are 1.5811 years and 1 year, respectively.

Exercise 10.9.6.

According to a YWCA Rape Crisis Center newsletter, 75% of rape victims know their attackers. A study is done to verify this.

Exercise 10.9.7. (Go to Solution)

According to a recent study, U.S. companies have an average maternity-leave of six weeks.

Exercise 10.9.8.

A recent drug survey showed an increase in use of drugs and alcohol among local high school students as compared to the national percent. Suppose that a survey of 100 local youths and 100 national youths is conducted to see if the percentage of drug and alcohol use is higher locally than nationally.

Exercise 10.9.9. (Go to Solution)

A new SAT study course is tested on 12 individuals. Pre-course and post-course scores are recorded. Of interest is the average increase in SAT scores.

Exercise 10.9.10.

University of Michigan researchers reported in the Journal of the National Cancer Institute that quitting smoking is especially beneficial for those under age 49. In this American Cancer Society study, the risk (probability) of dying of lung cancer was about the same as for those who had never smoked.

## For each problem below, fill in a hypothesis test solution sheet. The solution sheet is in the Appendix and can be copied. For the online version of the book, it is suggested that you copy the .doc or .pdf files.

### Note

If you are using a student-t distribution for a homework problem below, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

Exercise 10.9.11. (Go to Solution)

A powder diet is tested on 49 people and a liquid diet is tested on 36 different people. Of interest is whether the liquid diet yields a higher average weight loss than the powder diet. The powder diet group had an average weight loss of 42 pounds with a standard deviation of 12 pounds. The liquid diet group had an average weight loss of 45 pounds with a standard deviation of 14 pounds.

Exercise 10.9.12.

The average number of English courses taken in a two–year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The males took an average of 3 English courses with a standard deviation of 0.8. The females took an average of 4 English courses with a standard deviation of 1.0. Are the averages statistically the same?

Exercise 10.9.13. (Go to Solution)

A study is done to determine if students in the California state university system take longer to graduate than students enrolled in private universities. 100 students from both the California state university system and private universities are surveyed. Suppose that from years of research, it is known that the population standard deviations are 1.5811 years and 1 year, respectively. The following data are collected. The California state university system students took on average 4.5 years with a standard deviation of 0.8. The private university students took on average 4.1 years with a standard deviation of 0.3.

Exercise 10.9.14.

A new SAT study course is tested on 12 individuals. Pre-course and post-course scores are recorded. Of interest is the average increase in SAT scores. The following data is collected:

 Pre-course score Post-course score 1200 1300 960 920 1010 1100 840 880 1100 1070 1250 1320 860 860 1330 1370 790 770 990 1040 1110 1200 740 850

Exercise 10.9.15. (Go to Solution)

A recent drug survey showed an increase in use of drugs and alcohol among local high school seniors as compared to the national percent. Suppose that a survey of 100 local seniors and 100 national seniors is conducted to see if the percentage of drug and alcohol use is higher locally than nationally. Locally, 65 seniors reported using drugs or alcohol within the past month, while 60 national seniors reported using them.

Exercise 10.9.16.

A student at a four-year college claims that average enrollment at four–year colleges is higher than at two–year colleges in the United States. Two surveys are conducted. Of the 35 two–year colleges surveyed, the average enrollment was 5068 with a standard deviation of 4777. Of the 35 four-year colleges surveyed, the average enrollment was 5466 with a standard deviation of 8191. (Source: Microsoft Bookshelf)

Exercise 10.9.17. (Go to Solution)

A study was conducted by the U.S. Army to see if applying antiperspirant to soldiers’ feet for a few days before a major hike would help cut down on the number of blisters soldiers had on their feet. In the experiment, for three nights before they went on a 13-mile hike, a group of 328 West Point cadets put an alcohol-based antiperspirant on their feet. A “control group” of 339 soldiers put on a similar, but inactive, preparation on their feet. On the day of the hike, the temperature reached 83° F. At the end of the hike, 21% of the soldiers who had used the antiperspirant and 48% of the control group had developed foot blisters. Conduct a hypothesis test to see if the percent of soldiers using the antiperspirant was significantly lower than the control group. (Source: U.S. Army study reported in Journal of the American Academy of Dermatologists)

Exercise 10.9.18.

We are interested in whether the percents of female suicide victims for ages 15 to 24 are the same for the white and the black races in the United States. We randomly pick one year, 1992, to compare the races. The number of suicides estimated in the United States in 1992 for white females is 4930. 580 were aged 15 to 24. The estimate for black females is 330. 40 were aged 15 to 24. We will let female suicide victims be our population. (Source: the National Center for Health Statistics, U.S. Dept. of Health and Human Services)

Exercise 10.9.19. (Go to Solution)

At Rachel’s 11th birthday party, 8 girls were timed to see how long (in seconds) they could hold their breath in a relaxed position. After a two-minute rest, they timed themselves while jumping. The girls thought that the jumping would not affect their times, on average. Test their hypothesis.

 Relaxed time (seconds) Jumping time (seconds) 26 21 47 40 30 28 22 21 23 25 45 43 37 35 29 32

Exercise 10.9.20.

Elizabeth Mjelde, an art history professor, was interested in whether the value from the Golden Ratio formula, was the same in the Whitney Exhibit for works from 1900 – 1919 as for works from 1920 – 1942. 37 early works were sampled. They averaged 1.74 with a standard deviation of 0.11. 65 of the later works were sampled. They averaged 1.746 with a standard deviation of 0.1064. Do you think that there is a significant difference in the Golden Ratio calculation? (Source: data from Whitney Exhibit on loan to San Jose Museum of Art)

Exercise 10.9.21. (Go to Solution)

One of the questions in a study of marital satisfaction of dual–career couples was to rate the statement, “I’m pleased with the way we divide the responsibilities for childcare.” The ratings went from 1 (strongly agree) to 5 (strongly disagree). Below are ten of the paired responses for husbands and wives. Conduct a hypothesis test to see if the average difference in the husband’s versus the wife’s satisfaction level is negative (meaning that, within the partnership, the husband is happier than the wife).

 Wife’s score 2 2 3 3 4 2 1 1 2 4 Husband’s score 2 2 1 3 2 1 1 1 2 4

Exercise 10.9.22.

Ten individuals went on a low–fat diet for 12 weeks to lower their cholesterol. Evaluate the data below. Do you think that their cholesterol levels were significantly lowered?

 Starting cholesterol level Ending cholesterol level 140 140 220 230 110 120 240 220 200 190 180 150 190 200 360 300 280 300 260 240

Exercise 10.9.23. (Go to Solution)

Average entry level salaries for college graduates with mechanical engineering degrees and electrical engineering degrees are believed to be approximately the same. (Source: http:// www.graduatingengineer.com ). A recruiting office thinks that the average mechanical engineering salary is actually lower than the average electrical engineering salary. The recruiting office randomly surveys 50 entry level mechanical engineers and 60 entry level electrical engineers. Their average salaries were \$46,100 and \$46,700, respectively. Their standard deviations were \$3450 and \$4210, respectively. Conduct a hypothesis test to determine if you agree that the average entry level mechanical engineering salary is lower than the average entry level electrical engineering salary.

Exercise 10.9.24.

A recent year was randomly picked from 1985 to the present. In that year, there were 2051 Hispanic students at Cabrillo College out of a total of 12,328 students. At Lake Tahoe College, there were 321 Hispanic students out of a total of 2441 students. In general, do you think that the percent of Hispanic students at the two colleges is basically the same or different? (Source: Chancellor’s Office, California Community Colleges, November 1994)

Exercise 10.9.25. (Go to Solution)

Eight runners were convinced that the average difference in their individual times for running one mile versus race walking one mile was at most 2 minutes. Below are their times. Do you agree that the average difference is at most 2 minutes?

 Running time (minutes) Race walking time (minutes) 5.1 7.3 5.6 9.2 6.2 10.4 4.8 6.9 7.1 8.9 4.2 9.5 6.1 9.4 4.4 7.9

Exercise 10.9.26.

Marketing companies have collected data implying that teenage girls use more ring tones on their cellular phones than teenage boys do. In one particular study of 40 randomly chosen teenage girls and boys (20 of each) with cellular phones, the average number of ring tones for the girls was 3.2 with a standard deviation of 1.5. The average for the boys was 1.7 with a standard deviation of 0.8. Conduct a hypothesis test to determine if the averages are approximately the same or if the girls’ average is higher than the boys’ average.

Exercise 10.9.27. (Go to Solution)

While her husband spent 2½ hours picking out new speakers, a statistician decided to determine whether the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment. The population was Saturday afternoon shoppers. Out of 67 men, 24 said they enjoyed the activity. 8 of the 24 women surveyed claimed to enjoy the activity. Interpret the results of the survey.

Exercise 10.9.28.

We are interested in whether children’s educational computer software costs less, on average, than children’s entertainment software. 36 educational software titles were randomly picked from a catalog. The average cost was \$31.14 with a standard deviation of \$4.69. 35 entertainment software titles were randomly picked from the same catalog. The average cost was \$33.86 with a standard deviation of \$10.87. Decide whether children’s educational software costs less, on average, than children’s entertainment software. (Source: Educational Resources, December catalog)

Exercise 10.9.29. (Go to Solution)

Parents of teenage boys often complain that auto insurance costs more, on average, for teenage boys than for teenage girls. A group of concerned parents examines a random sample of insurance bills. The average annual cost for 36 teenage boys was \$679. For 23 teenage girls, it was \$559. From past years, it is known that the population standard deviation for each group is \$180. Determine whether or not you believe that the average cost for auto insurance for teenage boys is greater than that for teenage girls.

Exercise 10.9.30.

A group of transfer bound students wondered if they will spend the same average amount on texts and supplies each year at their four-year university as they have at their community college. They conducted a random survey of 54 students at their community college and 66 students at their local four-year university. The sample means were \$947 and \$1011, respectively. The population standard deviations are known to be \$254 and \$87, respectively. Conduct a hypothesis test to determine if the averages are statistically the same.

Exercise 10.9.31. (Go to Solution)

Joan Nguyen recently claimed that the proportion of college–age males with at least one pierced ear is as high as the proportion of college–age females. She conducted a survey in her classes. Out of 107 males, 20 had at least one pierced ear. Out of 92 females, 47 had at least one pierced ear. Do you believe that the proportion of males has reached the proportion of females?

Exercise 10.9.32.

Some manufacturers claim that non-hybrid sedan cars have a lower average miles per gallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get an average 31 mpg with a standard deviation of 7 mpg. Thirty-one non-hybrid sedans average 22 mpg with a standard deviation of 4 mpg. Suppose that the population standard deviations are known to be 6 and 3, respectively. Conduct a hypothesis test to the manufacturers claim.

Questions Exercise 10.9.33.Exercise 10.9.37. refer to the Terri Vogel’s data set (see Table of Contents).

Exercise 10.9.33. (Go to Solution)

Using the data from Lap 1 only, conduct a hypothesis test to determine if the average time for completing a lap in races is the same as it is in practices.

Exercise 10.9.34.

Repeat the test in Exercise 10.9.33., but use Lap 5 data this time.

Exercise 10.9.35. (Go to Solution)

Repeat the test in Exercise 10.9.33., but this time combine the data from Laps 1 and 5.

Exercise 10.9.36.

In 2 – 3 complete sentences, explain in detail how you might use Terri Vogel’s data to answer the following question. “Does Terri Vogel drive faster in races than she does in practices?”

Exercise 10.9.37. (Go to Solution)

Is the proportion of race laps Terri completes slower than 130 seconds less than the proportion of practice laps she completes slower than 135 seconds?

Exercise 10.9.38.

“To Breakfast or Not to Breakfast?” by Richard Ayore

In the American society, birthdays are one of those days that everyone looks forward to. People of different ages and peer groups gather to mark the 18th, 20th, … birthdays. During this time, one looks back to see what he or she had achieved for the past year, and also focuses ahead for more to come.

If, by any chance, I am invited to one of these parties, my experience is always different. Instead of dancing around with my friends while the music is booming, I get carried away by memories of my family back home in Kenya. I remember the good times I had with my brothers and sister while we did our daily routine.

Every morning, I remember we went to the shamba (garden) to weed our crops. I remember one day arguing with my brother as to why he always remained behind just to join us an hour later. In his defense, he said that he preferred waiting for breakfast before he came to weed. He said, “This is why I always work more hours than you guys!”

And so, to prove his wrong or right, we decided to give it a try. One day we went to work as usual without breakfast, and recorded the time we could work before getting tired and stopping. On the next day, we all ate breakfast before going to work. We recorded how long we worked again before getting tired and stopping. Of interest was our average increase in work time. Though not sure, my brother insisted that it is more than two hours. Using the data below, solve our problem.

 Work hours with breakfast Work hours without breakfast 8 6 7 5 9 5 5 4 9 7 8 7 10 7 7 5 6 6 9 5

## Try these multiple choice questions.

For questions Exercise 10.9.39.Exercise 10.9.40., use the following information.

A new AIDS prevention drugs was tried on a group of 224 HIV positive patients. Forty-five (45) patients developed AIDS after four years. In a control group of 224 HIV positive patients, 68 developed AIDS after four years. We want to test whether the method of treatment reduces the proportion of patients that develop AIDS after four years or if the proportions of the treated group and the untreated group stay the same.

Let the subscript t = treated patient and = untreated patient.

Exercise 10.9.39. (Go to Solution)

The appropriate hypotheses are:

 A. and B. and C. and D. and Exercise 10.9.40. (Go to Solution)

If the p -value is 0.0062 what is the conclusion (use α = 0.05  )?

 A. The method has no effect. B. The method reduces the proportion of HIV positive patients that develop AIDS after four years. C. The method increases the proportion of HIV positive patients that develop AIDS after four years. D. The test does not determine whether the method helps or does not help.

Exercise 10.9.41. (Go to Solution)

Lesley E. Tan investigated the relationship between left-handedness and right-handedness and motor competence in preschool children. Random samples of 41 left-handers and 41 right-handers were given several tests of motor skills to determine if there is evidence of a difference between the children based on this experiment. The experiment produced the means and standard deviations shown below. Determine the appropriate test and best distribution to use for that test.

 Left-handed Right-handed Sample size 41 41 Sample mean 97.5 98.1 Sample standard deviation 17.5 19.2

 A. Two independent means, normal distribution B. Two independent means, student-t distribution C. Matched or paired samples, student-t distribution D. Two population proportions, normal distribution

For questions Exercise 10.9.42.Exercise 10.9.43., use the following information.

An experiment is conducted to show that blood pressure can be consciously reduced in people trained in a “biofeedback exercise program.” Six (6) subjects were randomly selected and the blood pressure measurements were recorded before and after the training. The difference between blood pressures was calculated producing the following results: s d = 8.4. Using the data, test the hypothesis that the blood pressure has decreased after the training,

Exercise 10.9.42. (Go to Solution)

The distribution for the test is

 A. t 5 B. t 6 C. N ( − 10 . 2,8 . 4 ) D. Exercise 10.9.43. (Go to Solution)

If α = 0.05, the p -value and the conclusion are

 A. 0.0014; the blood pressure decreased after the training B. 0.0014; the blood pressure increased after the training C. 0.0155; the blood pressure decreased after the training D. 0.0155; the blood pressure increased after the training

For questions Exercise 10.9.44.Exercise 10.9.45., use the following information.

The Eastern and Western Major League Soccer conferences have a new Reserve Division that allows new players to develop their skills. As of May 25, 2005, the Reserve Division teams scored the following number of goals for 2005.

 Western Eastern Los Angeles 9 D.C. United 9 FC Dallas 3 Chicago 8 Chivas USA 4 Columbus 7 Real Salt Lake 3 New England 6 Colorado 4 MetroStars 5 San Jose 4 Kansas City 3

Conduct a hypothesis test to determine if the Western Reserve Division teams score, on average, fewer goals than the Eastern Reserve Division teams. Subscripts: 1 Western Reserve Division (W); 2 Eastern Reserve Division (E)

Exercise 10.9.44. (Go to Solution)

The exact distribution for the hypothesis test is:

 A. The normal distribution. B. The student-t distribution. C. The uniform distribution. D. The exponential distribution.

Exercise 10.9.45. (Go to Solution)

If the level of significance is 0.05, the conclusion is:

 A. The W Division teams score, on average, fewer goals than the E teams. B. The W Division teams score, on average, more goals than the E teams. C. The W teams score, on average, about the same number of goals as the E teams score. D. Unable to determine.

Questions Exercise 10.9.46.Exercise 10.9.48. refer to the following.

A researcher is interested in determining if a certain drug vaccine prevents West Nile disease. The vaccine with the drug is administered to 36 people and another 36 people are given a vaccine that does not contain the drug. Of the group that gets the vaccine with the drug, one (1) gets West Nile disease. Of the group that gets the vaccine without the drug, three (3) get West Nile disease. Conduct a hypothesis test to determine if the proportion of people that get the vaccine without the drug and get West Nile disease is more than the proportion of people that get the vaccine with the drug and get West Nile disease.

• “Drug” subscript: group who get the vaccine with the drug.

• “No Drug” subscript: group who get the vaccine without the drug

Exercise 10.9.46. (Go to Solution)

This is a test of:

 A. a test of two proportions B. a test of two independent means C. a test of a single mean D. a test of matched pairs.

Exercise 10.9.47. (Go to Solution)

An appropriate null hypothesis is:

 A. B. C. D. Exercise 10.9.48. (Go to Solution)

The p -value is 0.1517. At a 1% level of significance, the appropriate conclusion is

 A. the proportion of people that get the vaccine without the drug and get West Nile disease is less than the proportion of people that get the vaccine with the drug and get West Nile disease. B. the proportion of people that get the vaccine without the drug and get West Nile disease is more than the proportion of people that get the vaccine with the drug and get West Nile disease. C. the proportion of people that get the vaccine without the drug and get West Nile disease is more than or equal to the proportion of people that get the vaccine with the drug and get West Nile disease. D. the proportion of people that get the vaccine without the drug and get West Nile disease is no more than the proportion of people that get the vaccine with the drug and get West Nile disease.

Questions Exercise 10.9.49. and Exercise 10.9.50. refer to the following:

A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four (4) new students. She records their 18-holes scores before learning the technique and then after having taken her class. She conducts a hypothesis test. The data are as follows.

 Player 1 Player 2 Player 3 Player 4 Average score before class 83 78 93 87 Average score after class 80 80 86 86

Exercise 10.9.49. (Go to Solution)

This is a test of:

 A. a test of two independent means B. a test of two proportions C. a test of a single proportion D. a test of matched pairs.

Exercise 10.9.50. (Go to Solution)

The correct decision is:

 A. Reject H o B. Do not reject H o C. The test is inconclusive

Questions Exercise 10.9.51. and Exercise 10.9.52. refer to the following:

Suppose a statistics instructor believes that there is no significant difference between the average class scores of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples from each of the populations. The average and standard deviation for 35 statistics day students were 75.86 and 16.91. The average and standard deviation for 37 statistics night students were 75.41 and 19.73. The “day” subscript refers to the statistics day students. The “night” subscript refers to the statistics night students.

Exercise 10.9.51. (Go to Solution)

An appropriate alternate hypothesis for the hypothesis test is:

 A. μ day > μ night B. μ day < μ night C. μ day = μ night D. μ day ≠ μ night

Exercise 10.9.52. (Go to Solution)

A concluding statement is:

 A. The statistics night students average on Exam 2 is better than the statistics day students average on Exam 2. B. The statistics day students average on Exam 2 is better than the statistics night students average on Exam 2. C. There is no significant difference between the averages of the statistics day students and night students on Exam 2. D. There is a significant difference between the averages of the statistics day students and night students on Exam 2.

## Solutions to Exercises

A

B

A

D

C

 d. t 68 . 44 e. -1.04 f. 0.1519 h. Dec: do not reject null

Standard Normal

 e. z = 2 . 14 f. 0.0163 h. Decision: Reject null when α = 0 . 05 ; Do not reject null when α = 0 . 01

 e. 0.73 f. 0.2326 h. Decision: Do not reject null

 e. -7.33 f. 0 h. Decision: Reject null

 d. t 7 e. -1.51 f. 0.1755 h. Decision: Do not reject null

 d. t 9 e. t = − 1 . 86 f. 0.0479 h. Decision: Reject null, but run another test

 d. t 108 e. t = − 0 . 82 f. 0.2066 h. Decision: Do not reject null

 d. t 7 e. t = 2 . 9850 f. 0.0103 h. Decision: Reject null; The average difference is more than 2 minutes.

 e. 0.22 f. 0.4133 h. Decision: Do not reject null

 e. z = 2 . 50 f. 0.0063 h. Decision: Reject null

 e. -4.82 f. 0 h. Decision: Reject null

 d. t 20 . 32 e. -4.70 f. 0.0001 h. Decision: Reject null

 d. t 40 . 94 e. -5.08 f. 0 h. Decision: Reject null

 e. -0.95 f. 0.1705 h. Decision: Do not reject null

D

B

B

A

C

B

C

A

A

D

D

B

D

C

# 10.10. Review*

The next three questions refer to the following information:

In a survey at Kirkwood Ski Resort the following information was recorded:

 0 – 10 11 - 20 21 - 40 40+ Ski 10 12 30 8 Snowboard 6 17 12 5

Suppose that one person from of the above was randomly selected.

Exercise 10.10.1. (Go to Solution)

Find the probability that the person was a skier or was age 11 – 20.

Exercise 10.10.2. (Go to Solution)

Find the probability that the person was a snowboarder given he/she was age 21 – 40.

Exercise 10.10.3. (Go to Solution)

Explain which of the following are true and which are false.

 a. Sport and Age are independent events. b. Ski and age 11 – 20 are mutually exclusive events. c. d. Exercise 10.10.4. (Go to Solution)

The average length of time a person with a broken leg wears a cast is approximately 6 weeks. The standard deviation is about 3 weeks. Thirty people who had recently healed from broken legs were interviewed. State the distribution that most accurately reflects total time to heal for the thirty people.

Exercise 10.10.5. (Go to Solution)

The distribution for X is Uniform. What can we say for certain about the distribution for when n = 1?

 A. The distribution for is still Uniform with the same mean and standard dev. as the distribution for X . B. The distribution for is Normal with the different mean and a different standard deviation as the distribution for X . C. The distribution for is Normal with the same mean but a larger standard deviation than the distribution for X . D. The distribution for is Normal with the same mean but a smaller standard deviation than the distribution for X .

Exercise 10.10.6. (Go to Solution)

The distribution for X is uniform. What can we say for certain about the distribution for X when n = 50?

 A. The distribution for ∑X is still uniform with the same mean and standard deviation as the distribution for X . B. The distribution for ∑X is Normal with the same mean but a larger standard deviation as the distribution for X . C. The distribution for ∑X is Normal with a larger mean and a larger standard deviation than the distribution for X . D. The distribution for ∑X is Normal with the same mean but a smaller standard deviation than the distribution for X .

The next three questions refer to the following information:

A group of students measured the lengths of all the carrots in a five-pound bag of baby carrots. They calculated the average length of baby carrots to be 2.0 inches with a standard deviation of 0.25 inches. Suppose we randomly survey 16 five-pound bags of baby carrots.

Exercise 10.10.7. (Go to Solution)

State the approximate distribution for , the distribution for the average lengths of baby carrots in 16 five-pound bags. Exercise 10.10.8.

Explain why we cannot find the probability that one individual randomly chosen carrot is greater than 2.25 inches.

Exercise 10.10.9. (Go to Solution)

Find the probability that is between 2 and 2.25 inches.

The next three questions refer to the following information:

At the beginning of the term, the amount of time a student waits in line at the campus store is normally distributed with a mean of 5 minutes and a standard deviation of 2 minutes.

Exercise 10.10.10. (Go to Solution)

Find the 90th percentile of waiting time in minutes.

Exercise 10.10.11. (Go to Solution)

Find the median waiting time for one student.

Exercise 10.10.12. (Go to Solution)

Find the probability that the average waiting time for 40 students is at least 4.5 minutes.

## Solutions to Exercises  a. False b. False c. True d. False

N ( 180 , 16 . 43 )

A

C 0.5000

7.6

5

0.9431

# 10.11. Lab: Hypothesis Testing for Two Means and Two Proportions*

Class Time:

Names:

## Student Learning Outcomes:

• The student will select the appropriate distributions to use in each case.

• The student will conduct hypothesis tests and interpret the results.

## Supplies:

• The business section from two consecutive days’ newspapers

• 3 small packages of M&Ms®

• 5 small packages of Reese’s Pieces®

## Increasing Stocks Survey

Look at yesterday’s newspaper business section. Conduct a hypothesis test to determine if the proportion of New York Stock Exchange (NYSE) stocks that increased is greater than the proportion of NASDAQ stocks that increased. As randomly as possible, choose 40 NYSE stocks and 32 NASDAQ stocks and complete the following statements.

1. H o

2. H a

3. In words, define the Random Variable. ____________=

4. The distribution to use for the test is:

5. Calculate the test statistic using your data.

6. Draw a graph and label it appropriately. Shade the actual level of significance.

 a. Graph: Figure 10.8. b. Calculate the p-value:

7. Do you reject or not reject the null hypothesis? Why?

8. Write a clear conclusion using a complete sentence.

## Decreasing Stocks Survey

Randomly pick 8 stocks from the newspaper. Using two consecutive days’ business sections, test whether the stocks went down, on average, for the second day.

1. H o

2. H a

3. In words, define the Random Variable. ____________=

4. The distribution to use for the test is:

5. Calculate the test statistic using your data.

6. Draw a graph and label it appropriately. Shade the actual level of significance.

 a. Graph: Figure 10.9. b. Calculate the p-value:

7. Do you reject or not reject the null hypothesis? Why?

8. Write a clear conclusion using a complete sentence.

## Candy Survey

Buy three small packages of M&Ms and 5 small packages of Reese’s Pieces (same net weight as the M&Ms). Test whether or not the average number of candy pieces per package is the same for the two brands.

1. H o :

2. H a :

3. In words, define the random variable. __________=

4. What distribution should be used for this test?

5. Calculate the test statistic using your data.

6. Draw a graph and label it appropriately. Shade the actual level of significance.

 a. Graph: Figure 10.10. b. Calculate the p-value:

7. Do you reject or not reject the null hypothesis? Why?

8. Write a clear conclusion using a complete sentence.

## Shoe Survey

Test whether women have, on average, more pairs of shoes than men. Include all forms of sneakers, shoes, sandals, and boots. Use your class as the sample.

1. H o

2. H a

3. In words, define the Random Variable. ____________=

4. The distribution to use for the test is:

5. Calculate the test statistic using your data.

6. Draw a graph and label it appropriately. Shade the actual level of significance.

 a. Graph: Figure 10.11. b. Calculate the p-value:

7. Do you reject or not reject the null hypothesis? Why?

8. Write a clear conclusion using a complete sentence.