# 10.E: Hypothesis Testing with Two Samples (Exercises)

These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.

## 10.2: Two Population Means with Unknown Standard Deviations

Use the following information to answer the next 15 exercises: Indicate if the hypothesis test is for

1. independent group means, population standard deviations, and/or variances known
2. independent group means, population standard deviations, and/or variances unknown
3. matched or paired samples
4. single mean
5. two proportions
6. single proportion

Exercise 10.2.3

It is believed that 70% of males pass their drivers test in the first attempt, while 65% of females pass the test in the first attempt. Of interest is whether the proportions are in fact equal.

two proportions

Exercise 10.2.4

A new laundry detergent is tested on consumers. Of interest is the proportion of consumers who prefer the new brand over the leading competitor. A study is done to test this.

Exercise 10.2.5

A new windshield treatment claims to repel water more effectively. Ten windshields are tested by simulating rain without the new treatment. The same windshields are then treated, and the experiment is run again. A hypothesis test is conducted.

matched or paired samples

Exercise 10.2.6

### S 10.3.3

Subscripts: 1 = boys, 2 = girls

1. $$H_{0}: \mu_{1} \leq \mu_{2}$$
2. $$H_{a}: \mu_{1} > \mu_{2}$$
3. The random variable is the difference in the mean auto insurance costs for boys and girls.
4. normal
5. test statistic: $$z = 2.50$$
6. $$p\text{-value}: 0.0062$$
7. Check student’s solution.
1. $$\alpha: 0.05$$
2. Decision: Reject the null hypothesis.
3. Reason for Decision: $$p\text{-value} < \alpha$$
4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean cost of auto insurance for teenage boys is greater than that for girls.

### Q 10.3.4

A group of transfer bound students wondered if they will spend the same mean 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$1,011, respectively. The population standard deviations are known to be $254 and$87, respectively. Conduct a hypothesis test to determine if the means are statistically the same.

### Q 10.3.5

Some manufacturers claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a mean of 31 mpg with a standard deviation of seven mpg. Thirty-one non-hybrid sedans get a mean of 22 mpg with a standard deviation of four mpg. Suppose that the population standard deviations are known to be six and three, respectively. Conduct a hypothesis test to evaluate the manufacturers claim.

### S 10.3.5

Subscripts: 1 = non-hybrid sedans, 2 = hybrid sedans

1. $$H_{0}: \mu_{1} \geq \mu_{2}$$
2. $$H_{a}: \mu_{1} < \mu_{2}$$
3. The random variable is the difference in the mean miles per gallon of non-hybrid sedans and hybrid sedans.
4. normal
5. test statistic: 6.36
6. $$p\text{-value}: 0$$
7. Check student’s solution.
1. $$\alpha: 0.05$$
2. Decision: Reject the null hypothesis.
3. Reason for decision: $$p\text{-value} < \alpha$$
4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean miles per gallon of non-hybrid sedans is less than that of hybrid sedans.

### Q 10.3.6

A baseball fan wanted to know if there is a difference between the number of games played in a World Series when the American League won the series versus when the National League won the series. From 1922 to 2012, the population standard deviation of games won by the American League was 1.14, and the population standard deviation of games won by the National League was 1.11. Of 19 randomly selected World Series games won by the American League, the mean number of games won was 5.76. The mean number of 17 randomly selected games won by the National League was 5.42. Conduct a hypothesis test.

### Q 10.3.7

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 one (strongly agree) to five (strongly disagree). Table contains ten of the paired responses for husbands and wives. Conduct a hypothesis test to see if the mean 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

### S 10.3.7

1. $$H_{0}: \mu_{d} = 0$$
2. $$H_{a}: \mu_{d} < 0$$
3. The random variable $$X_{d}$$ is the average difference between husband’s and wife’s satisfaction level.
4. $$t_{9}$$
5. test statistic: $$t = –1.86$$
6. $$p\text{-value}: 0.0479$$
7. Check student’s solution
1. $$\alpha: 0.05$$
2. Decision: Reject the null hypothesis, but run another test.
3. Reason for Decision: $$p\text{-value} < \alpha$$
4. Conclusion: This is a weak test because alpha and the p-value are close. However, there is insufficient evidence to conclude that the mean difference is negative.

## 10.4: Comparing Two Independent Population Proportions

Use the following information for the next five exercises. Two types of phone operating system are being tested to determine if there is a difference in the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones with OS1 had system failures within the first eight hours of operation. Nine out of another random sample of 150 phones with OS2 had system failures within the first eight hours of operation. OS2 is believed to be more stable (have fewer crashes) than OS1.

Exercise 10.4.2

Is this a test of means or proportions?

Exercise 10.4.3

What is the random variable?

$$P'_{OS_{1}} - P'_{OS_{2}} =$$ difference in the proportions of phones that had system failures within the first eight hours of operation with OS1 and OS2.

Exercise 10.4.4

State the null and alternative hypotheses.

Exercise 10.4.5

What is the $$p\text{-value}$$?

0.1018

Exercise 10.4.6

What can you conclude about the two operating systems?

Use the following information to answer the next twelve exercises. In the recent Census, three percent of the U.S. population reported being of two or more races. However, the percent varies tremendously from state to state. Suppose that two random surveys are conducted. In the first random survey, out of 1,000 North Dakotans, only nine 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.

Exercise 10.4.7

Is this a test of means or proportions?

proportions

Exercise 10.4.8

State the null and alternative hypotheses.

1. $$H_{0}$$: _________
2. $$H_{a}$$: _________

Exercise 10.4.9

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

right-tailed

Exercise 10.4.10

What is the random variable of interest for this test?

Exercise 10.4.11

In words, define the random variable for this test.

The random variable is the difference in proportions (percents) of the populations that are of two or more races in Nevada and North Dakota.

Exercise 10.4.12

Which distribution (normal or Student's t) would you use for this hypothesis test?

Exercise 10.4.13

Explain why you chose the distribution you did for the Exercise 10.56.

Our sample sizes are much greater than five each, so we use the normal for two proportions distribution for this hypothesis test.

Exercise 10.4.14

Calculate the test statistic.

Exercise 10.4.15

Sketch a graph of the situation. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the $$p\text{-value}$$.

Check student’s solution.

Exercise 10.4.16

Find the $$p\text{-value}$$.

Exercise 10.4.17

At a pre-conceived $$\alpha = 0.05$$, what is your:

1. Decision:
2. Reason for the decision:
3. Conclusion (write out in a complete sentence):

1. Reject the null hypothesis.
2. $$p\text{-value} < \alpha$$
3. At the 5% significance level, there is sufficient evidence to conclude that the proportion (percent) of the population that is of two or more races in Nevada is statistically higher than that in North Dakota.

Exercise 10.4.18

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?

DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in [link]. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

NOTE

If you are using a Student's t-distribution for one of the following homework problems, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.)

### Q 10.4.1

A recent drug survey showed an increase in the 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 proportion 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.

### Q 10.4.2

We are interested in whether the proportions of female suicide victims for ages 15 to 24 are the same for the whites and the blacks 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 4,930. Five hundred eighty were aged 15 to 24. The estimate for black females is 330. Forty were aged 15 to 24. We will let female suicide victims be our population.

### S 10.4.2

1. $$H_{0}: P_{W} = P_{B}$$
2. $$H_{a}: P_{W} \neq P_{B}$$
3. The random variable is the difference in the proportions of white and black suicide victims, aged 15 to 24.
4. normal for two proportions
5. test statistic: –0.1944
6. $$p\text{-value}: 0.8458$$
7. Check student’s solution.
1. $$\alpha: 0.05$$
2. Decision: Reject the null hypothesis.
3. Reason for decision: $$p\text{-value} > \alpha$$
4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the proportions of white and black female suicide victims, aged 15 to 24, are different.

### Q 10.4.3

Elizabeth Mjelde, an art history professor, was interested in whether the value from the Golden Ratio formula, $$\left(\frac{(larger + smaller dimension}{larger dimension}\right$$ was the same in the Whitney Exhibit for works from 1900 to 1919 as for works from 1920 to 1942. Thirty-seven early works were sampled, averaging 1.74 with a standard deviation of 0.11. Sixty-five of the later works were sampled, averaging 1.746 with a standard deviation of 0.1064. Do you think that there is a significant difference in the Golden Ratio calculation?

### Q 10.4.4

A recent year was randomly picked from 1985 to the present. In that year, there were 2,051 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 2,441 students. In general, do you think that the percent of Hispanic students at the two colleges is basically the same or different?

### S 10.4.4

Subscripts: 1 = Cabrillo College, 2 = Lake Tahoe College

1. $$H_{0}: p_{1} = p_{2}$$
2. $$H_{a}: p_{1} \neq p_{2}$$
3. The random variable is the difference between the proportions of Hispanic students at Cabrillo College and Lake Tahoe College.
4. normal for two proportions
5. test statistic: 4.29
6. $$p\text{-value}: 0.00002$$
7. Check student’s solution.
1. $$\alpha: 0.05$$
2. Decision: Reject the null hypothesis.
3. Reason for decision: p-value < alpha
4. Conclusion: There is sufficient evidence to conclude that the proportions of Hispanic students at Cabrillo College and Lake Tahoe College are different.

Use the following information to answer the next three exercises. Neuroinvasive West Nile virus is a severe disease that affects a person’s nervous system . It is spread by the Culex species of mosquito. In the United States in 2010 there were 629 reported cases of neuroinvasive West Nile virus out of a total of 1,021 reported cases and there were 486 neuroinvasive reported cases out of a total of 712 cases reported in 2011. Is the 2011 proportion of neuroinvasive West Nile virus cases more than the 2010 proportion of neuroinvasive West Nile virus cases? Using a 1% level of significance, conduct an appropriate hypothesis test.

• “2011” subscript: 2011 group.
• “2010” subscript: 2010 group

### Q 10.4.5

This is:

1. a test of two proportions
2. a test of two independent means
3. a test of a single mean
4. a test of matched pairs.

### Q 10.4.6

An appropriate null hypothesis is:

1. $$p_{2011} \leq p_{2010}$$
2. $$p_{2011} \geq p_{2010}$$
3. $$\mu_{2011} \leq \mu_{2010}$$
4. $$p_{2011} > p_{2010}$$

a

### Q 10.4.7

The $$p\text{-value}$$ is 0.0022. At a 1% level of significance, the appropriate conclusion is

1. There is sufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is less than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease.
2. There is insufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is more than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease.
3. There is insufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is less than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease.
4. There is sufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is more than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease.

### Q 10.4.8

Researchers conducted a study to find out if there is a difference in the use of eReaders by different age groups. Randomly selected participants were divided into two age groups. In the 16- to 29-year-old group, 7% of the 628 surveyed use eReaders, while 11% of the 2,309 participants 30 years old and older use eReaders.

### S 10.4.9

Test: two independent sample proportions.

Random variable: $$p′_{1} - p′_{2}$$

Distribution:

$$H_{0}: p_{1} = p_{2}$$

$$H_{a}: p_{1} \neq p_{2}$$

The proportion of eReader users is different for the 16- to 29-year-old users from that of the 30 and older users.

Graph: two-tailed

$$p\text{-value}: 0.0033$$

Decision: Reject the null hypothesis.

Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that the proportion of eReader users 16 to 29 years old is different from the proportion of eReader users 30 and older.

### Q 10.4.10

are considered obese if their body mass index (BMI) is at least 30. The researchers wanted to determine if the proportion of women who are obese in the south is less than the proportion of southern men who are obese. The results are shown in Table. Test at the 1% level of significance.

Number who are obese Sample size
Men 42,769 155,525
Women 67,169 248,775

### Q 10.4.11

Two computer users were discussing tablet computers. A higher proportion of people ages 16 to 29 use tablets than the proportion of people age 30 and older. Table details the number of tablet owners for each age group. Test at the 1% level of significance.

16–29 year olds 30 years old and older
Own a Tablet 69 231
Sample Size 628 2,309

### S 10.4.11

Test: two independent sample proportions

Random variable: $$p′_{1} - p′_{2}$$

Distribution:

$$H_{0}: p_{1} = p_{2}$$

$$H_{a}: p_{1} > p_{2}$$

A higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older.

Graph: right-tailed

$$p\text{-value}: 0.2354$$

Decision: Do not reject the $$H_{0}$$.

Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older.

### Q 10.4.12

A group of friends debated whether more men use smartphones than women. They consulted a research study of smartphone use among adults. The results of the survey indicate that of the 973 men randomly sampled, 379 use smartphones. For women, 404 of the 1,304 who were randomly sampled use smartphones. Test at the 5% level of significance.

### Q 10.4.13

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. Eight of the 24 women surveyed claimed to enjoy the activity. Interpret the results of the survey.

### S 10.4.13

Subscripts: 1: men; 2: women

1. $$H_{0}: p_{1} \leq p_{2}$$
2. $$H_{a}: p_{1} > p_{2}$$
3. $$P'_{1} - P\_{2}$$ is the difference between the proportions of men and women who enjoy shopping for electronic equipment.
4. normal for two proportions
5. test statistic: 0.22
6. $$p\text{-value}: 0.4133$$
7. Check student’s solution.
1. $$\alpha: 0.05$$
2. Decision: Do not reject the null hypothesis.
3. Reason for Decision: $$p\text{-value} > \alpha$$
4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the proportion of men who enjoy shopping for electronic equipment is more than the proportion of women.

### Q 10.4.14

We are interested in whether children’s educational computer software costs less, on average, than children’s entertainment software. Thirty-six educational software titles were randomly picked from a catalog. The mean cost was $31.14 with a standard deviation of$4.69. Thirty-five entertainment software titles were randomly picked from the same catalog. The mean 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.

### Q 10.4.15

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?

### S 10.4.15

1. $$H_{0}: p_{1} = p_{2}$$
2. $$H_{a}: p_{1} \neq p_{2}$$
3. $$P'_{1} - P\_{2}$$ is the difference between the proportions of men and women that have at least one pierced ear.
4. normal for two proportions
5. test statistic: –4.82
6. $$p\text{-value}: 0$$
7. Check student’s solution.
1. $$\alpha: 0.05$$
2. Decision: Reject the null hypothesis.
3. Reason for Decision: $$p\text{-value} < \alpha$$
4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportions of males and females with at least one pierced ear is different.

### Q 10.4.16

Use the data sets found in [link] to answer this exercise. 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?

### Q 10.4.17

"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 has 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 him 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 mean increase in work time. Though not sure, my brother insisted that it was more than two hours. Using the data in Table, 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

### S 10.4.17

1. $$H_{0}: \mu_{d} = 0$$
2. $$H_{a}: \mu_{d} > 0$$
3. The random variable $$X_{d}$$ is the mean difference in work times on days when eating breakfast and on days when not eating breakfast.
4. $$t_{9}$$
5. test statistic: 4.8963
6. $$p\text{-value}: 0.0004$$
7. Check student’s solution.
1. $$\alpha: 0.05$$
2. Decision: Reject the null hypothesis.
3. Reason for Decision:$$p\text{-value} < \alpha$$
4. Conclusion: At the 5% level of significance, there is sufficient evidence to conclude that the mean difference in work times on days when eating breakfast and on days when not eating breakfast has increased.

## 10.5: Matched or Paired Samples

Use the following information to answer the next five exercises. A study was conducted to test the effectiveness of a software patch in reducing system failures over a six-month period. Results for randomly selected installations are shown in Table. The “before” value is matched to an “after” value, and the differences are calculated. The differences have a normal distribution. Test at the 1% significance level.

Installation A B C D E F G H
Before 3 6 4 2 5 8 2 6
After 1 5 2 0 1 0 2 2

Exercise 10.5.4

What is the random variable?

the mean difference of the system failures

Exercise 10.5.5

State the null and alternative hypotheses.

Exercise 10.5.6

What is the $$p\text{-value}$$?

0.0067

Exercise 10.5.7

Draw the graph of the $$p\text{-value}$$.

Exercise 10.5.8

What conclusion can you draw about the software patch?

With a $$p\text{-value} 0.0067$$, we can reject the null hypothesis. There is enough evidence to support that the software patch is effective in reducing the number of system failures.

Use the following information to answer next five exercises. A study was conducted to test the effectiveness of a juggling class. Before the class started, six subjects juggled as many balls as they could at once. After the class, the same six subjects juggled as many balls as they could. The differences in the number of balls are calculated. The differences have a normal distribution. Test at the 1% significance level.

Subject A B C D E F
Before 3 4 3 2 4 5
After 4 5 6 4 5 7

Exercise 10.5.9

State the null and alternative hypotheses.

Exercise 10.5.10

What is the $$p\text{-value}$$?

0.0021

Exercise 10.5.11

What is the sample mean difference?

Exercise 10.5.12

Draw the graph of the $$p\text{-value}$$.

Exercise 10.5.13

What conclusion can you draw about the juggling class?

Use the following information to answer the next five exercises. A doctor wants to know if a blood pressure medication is effective. Six subjects have their blood pressures recorded. After twelve weeks on the medication, the same six subjects have their blood pressure recorded again. For this test, only systolic pressure is of concern. Test at the 1% significance level.

Patient A B C D E F
Before 161 162 165 162 166 171
After 158 159 166 160 167 169

Exercise 10.5.14

State the null and alternative hypotheses.

$$H_{0}: \mu_{d} \geq 0$$

$$H_{a}: \mu_{d} < 0$$

Exercise 10.5.15

What is the test statistic?

Exercise 10.5.16

What is the $$p\text{-value}$$?

0.0699

Exercise 10.5.17

What is the sample mean difference?

Exercise 10.5.18

What is the conclusion?

We decline to reject the null hypothesis. There is not sufficient evidence to support that the medication is effective.

### Bringing It Together

Use the following information to answer the next ten exercises. indicate which of the following choices best identifies the hypothesis test.

1. independent group means, population standard deviations and/or variances known
2. independent group means, population standard deviations and/or variances unknown
3. matched or paired samples
4. single mean
5. two proportions
6. single proportion

Exercise 10.5.19

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

Exercise 10.5.20

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

e

Exercise 10.5.21

The mean 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 nine males and 16 females.

Exercise 10.5.22

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

d

Exercise 10.5.23

A study is done to determine if students in the California state university system take longer to graduate than students enrolled in private universities. One hundred 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 one year, respectively.

Exercise 10.5.24

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

f

Exercise 10.5.25

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

Exercise 10.5.26

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 proportion of drug and alcohol use is higher locally than nationally.

e

Exercise 10.5.27

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

Pre-course score Post-course score
1 300
960 920
1010 1100
840 880
1100 1070
1250 1320
860 860
1330 1370
790 770
990 1040
1110 1200
740 850

Exercise 10.5.28

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.

f

Exercise 10.5.29

Lesley E. Tan investigated the relationship between left-handedness vs. right-handedness and motor competence in preschool children. Random samples of 41 left-handed preschool children and 41 right-handed preschool children 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 Table. 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
1. Two independent means, normal distribution
2. Two independent means, Student’s-t distribution
3. Matched or paired samples, Student’s-t distribution
4. Two population proportions, normal distribution

Exercise 10.5.30

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-hole scores before learning the technique and then after having taken her class. She conducts a hypothesis test. The data are as Table.

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

This is:

1. a test of two independent means.
2. a test of two proportions.
3. a test of a single mean.
4. a test of a single proportion.

a

DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

NOTE

If you are using a Student's t-distribution for the homework problems, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.)

### Q 10.5.1

Ten individuals went on a low–fat diet for 12 weeks to lower their cholesterol. The data are recorded in Table. 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

### S 10.5.1

$$p\text{-value} = 0.1494$$

At the 5% significance level, there is insufficient evidence to conclude that the medication lowered cholesterol levels after 12 weeks.

Use the following information to answer the next two exercises. A new AIDS prevention drug was tried on a group of 224 HIV positive patients. Forty-five 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 $$ut =$$ untreated patient.

### Q 10.5.2

The appropriate hypotheses are:

1. $$H_{0}: p_{t} < p_{ut}$$ and $$H_{a}: p_{t} \geq p_{ut}$$
2. $$H_{0}: p_{t} \leq p_{ut}$$ and $$H_{a}: p_{t} > p_{ut}$$
3. $$H_{0}: p_{t} = p_{ut}$$ and $$H_{a}: p_{t} \neq p_{ut}$$
4. $$H_{0}: p_{t} = p_{ut}$$ and $$H_{a}: p_{t} < p_{ut}$$

### Q 10.5.3

If the $$p\text{-value}$$ is 0.0062 what is the conclusion (use $$\alpha = 0.05$$)?

1. The method has no effect.
2. There is sufficient evidence to conclude that the method reduces the proportion of HIV positive patients who develop AIDS after four years.
3. There is sufficient evidence to conclude that the method increases the proportion of HIV positive patients who develop AIDS after four years.
4. There is insufficient evidence to conclude that the method reduces the proportion of HIV positive patients who develop AIDS after four years.

### S 10.5.3

b

Use the following information to answer the next two exercises. An experiment is conducted to show that blood pressure can be consciously reduced in people trained in a “biofeedback exercise program.” Six subjects were randomly selected and blood pressure measurements were recorded before and after the training. The difference between blood pressures was calculated (after - before) producing the following results: $$\bar{x}_{d} = -10.2$$ $$s_{d} = 8.4$$. Using the data, test the hypothesis that the blood pressure has decreased after the training.

### Q 10.5.4

The distribution for the test is:

1. $$t_{5}$$
2. $$t_{6}$$
3. $$N(-10.2, 8.4)$$
4. $$N\left(-10.2, \frac{8.4}{\sqrt{6}}\right)$$

### Q 10.5.5

If $$\alpha = 0.05$$, the $$p\text{-value}$$ and the conclusion are

1. 0.0014; There is sufficient evidence to conclude that the blood pressure decreased after the training.
2. 0.0014; There is sufficient evidence to conclude that the blood pressure increased after the training.
3. 0.0155; There is sufficient evidence to conclude that the blood pressure decreased after the training.
4. 0.0155; There is sufficient evidence to conclude that the blood pressure increased after the training.

c

### Q 10.5.6

A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four new students. She records their 18-hole 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
Mean score before class 83 78 93 87
Mean score after class 80 80 86 86

The correct decision is:

1. Reject $$H_{0}$$.
2. Do not reject the $$H_{0}$$.

### Q 10.5.7

A local cancer support group believes that the estimate for new female breast cancer cases in the south is higher in 2013 than in 2012. The group compared the estimates of new female breast cancer cases by southern state in 2012 and in 2013. The results are in Table.

Southern States 2012 2013
Alabama 3,450 3,720
Arkansas 2,150 2,280
Florida 15,540 15,710
Georgia 6,970 7,310
Kentucky 3,160 3,300
Louisiana 3,320 3,630
Mississippi 1,990 2,080
North Carolina 7,090 7,430
Oklahoma 2,630 2,690
South Carolina 3,570 3,580
Tennessee 4,680 5,070
Texas 15,050 14,980
Virginia 6,190 6,280

### S 10.5.7

Test: two matched pairs or paired samples (t-test)

Random variable: $$\bar{X}_{d}$$

Distribution: $$t_{12}$$

$$H_{0}: \mu_{d} = 0 H_{a}: \mu_{d} > 0$$

The mean of the differences of new female breast cancer cases in the south between 2013 and 2012 is greater than zero. The estimate for new female breast cancer cases in the south is higher in 2013 than in 2012.

Graph: right-tailed

$$p\text{-value}: 0.0004$$

Decision: Reject $$H_{0}$$

Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that there was a higher estimate of new female breast cancer cases in 2013 than in 2012.

### Q 10.5.8

A traveler wanted to know if the prices of hotels are different in the ten cities that he visits the most often. The list of the cities with the corresponding hotel prices for his two favorite hotel chains is in Table. Test at the 1% level of significance.

Cities Hyatt Regency prices in dollars Hilton prices in dollars
Atlanta 107 169
Boston 358 289
Chicago 209 299
Dallas 209 198
Denver 167 169
Indianapolis 179 214
Los Angeles 179 169
New York City 625 459
Washington, DC 245 239

### Q 10.5.9

A politician asked his staff to determine whether the underemployment rate in the northeast decreased from 2011 to 2012. The results are in Table.

Northeastern States 2011 2012
Connecticut 17.3 16.4
Delaware 17.4 13.7
Maine 19.3 16.1
Maryland 16.0 15.5
Massachusetts 17.6 18.2
New Hampshire 15.4 13.5
New Jersey 19.2 18.7
New York 18.5 18.7
Ohio 18.2 18.8
Pennsylvania 16.5 16.9
Rhode Island 20.7 22.4
Vermont 14.7 12.3
West Virginia 15.5 17.3

### S 10.5.9

Test: matched or paired samples (t-test)

Difference data: $$\{–0.9, –3.7, –3.2, –0.5, 0.6, –1.9, –0.5, 0.2, 0.6, 0.4, 1.7, –2.4, 1.8\}$$

Random Variable: $$\bar{X}_{d}$$

Distribution: $$H_{0}: \mu_{d} = 0 H_{a}: \mu_{d} < 0$$

The mean of the differences of the rate of underemployment in the northeastern states between 2012 and 2011 is less than zero. The underemployment rate went down from 2011 to 2012.

Graph: left-tailed.

$$p\text{-value}: 0.1207$$
Decision: Do not reject $$H_{0}$$.