Content check — Statistics, Chapter 10: Hypothesis Testing with Two Samples

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

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

Answer

two proportions

10-3 §10.1 source ↗

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

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.

Answer

matched or paired samples

10-5 §10.1 source ↗

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

The average worker in Germany gets eight weeks of paid vacation.

Answer

single mean

10-7 §10.1 source ↗

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

It is believed that the average grade on an English essay in a particular school system is higher for females than for males. A random sample of 31 females had a mean score of 82 with a standard deviation of 3, and a random sample of 25 males had a mean score of 76 with a standard deviation of 4.

Answer

independent group means, population standard deviations and/or variances unknown

10-9 §10.1 source ↗

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

In a random sample of 100 forests in the United States, 56 were coniferous or contained conifers. In a random sample of 80 forests in Mexico, 40 were coniferous or contained conifers. Is the proportion of conifers in the United States statistically more than the proportion of conifers in Mexico?

Answer

two proportions

10-11 §10.1 source ↗

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

It is thought that teenagers sleep more than adults on average. A study is done to verify this. A sample of 16 teenagers has a mean of 8.9 hours slept and a standard deviation of 1.2. A sample of 12 adults has a mean of 6.9 hours slept and a standard deviation of 0.6.

Answer

independent group means, population standard deviations and/or variances unknown

10-13 §10.1 source ↗

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

A sample of 12 in-state graduate school programs at School A has a mean tuition of $64,000 with a standard deviation of $8,000. At School B, a sample of 16 in-state graduate programs has a mean tuition of $80,000 with a standard deviation of $6,000. On average, are the mean tuitions different?

Answer

independent group means, population standard deviations and/or variances unknown

10-15 §10.1 source ↗

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

A high school principal claims that 30 percent of student athletes drive themselves to school, while 4 percent of nonathletes drive themselves to school. In a sample of 20 student athletes, 45 percent drive themselves to school. In a sample of 35 nonathlete students, 6 percent drive themselves to school. Is the percent of student athletes who drive themselves to school more than the percent of nonathletes?

Answer

two proportions

10-17 §10.1 source ↗

Use the following information to answer the next three exercises: A study is done to determine which of two soft drinks has more sugar. There are 13 cans of Beverage A in a sample and six cans of Beverage B. The mean amount of sugar in Beverage A is 36 grams with a standard deviation of 0.6 grams. The mean amount of sugar in Beverage B is 38 grams with a standard deviation of 0.8 grams. The researchers believe that Beverage B has more sugar than Beverage A, on average. Both populations have normal distributions.

What is the random variable?

Answer

The random variable is the difference between the mean amounts of sugar in the two soft drinks.

10-19 §10.1 source ↗

Use the following information to answer the next 12 exercises. The U.S. Centers for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites.

Is this a test of means or proportions?

Answer

means

10-21 §10.1 source ↗

Use the following information to answer the next 12 exercises. The U.S. Centers for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites.

Is this a right-tailed, left-tailed, or two-tailed test?

Answer

two-tailed

10-23 §10.1 source ↗

Use the following information to answer the next 12 exercises. The U.S. Centers for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites.

In words, define the random variable of interest for this test.

Answer

the difference between the mean life spans of whites and nonwhites

10-25 §10.1 source ↗

Use the following information to answer the next 12 exercises. The U.S. Centers for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites.

Explain why you chose the distribution you did for Exercise 10.24.

Answer

This is a comparison of two population means with unknown population standard deviations.

10-27 §10.1 source ↗

Use the following information to answer the next 12 exercises. The U.S. Centers for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites.

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

Answer

Check student’s solution.

10-29 §10.1 source ↗

Use the following information to answer the next 12 exercises. The U.S. Centers for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites.

At a preconceived α = 0.05, write the following:

a. Your decision:

b. The reason for your decision:

c. Your conclusion (write out in a complete sentence):

Answer

a. Reject the null hypothesis.

b. p-value < 0.05

c. There is not enough evidence at the 5 percent level of significance to support the claim that life expectancy in the 1900s is different between whites and nonwhites.

10-31 §10.2 source ↗

Use the following information to answer the next five exercises. The mean speeds of fastball pitches from two different baseball pitchers are to be compared. A sample of 14 fastball pitches is measured from each pitcher. The populations have normal distributions. Table 10.18 shows the result. Scouters believe that Rodriguez pitches a speedier fastball.

What is the random variable?

Table 10.18 in the book ↗

Pitcher	Sample Mean Speed of Pitches (mph)	Population Standard Deviation
Wesley	86	3
Rodriguez	91	7

Answer

the difference in mean speeds of the fastball pitches of the two pitchers

10-33 §10.2 source ↗

Use the following information to answer the next five exercises. The mean speeds of fastball pitches from two different baseball pitchers are to be compared. A sample of 14 fastball pitches is measured from each pitcher. The populations have normal distributions. Table 10.18 shows the result. Scouters believe that Rodriguez pitches a speedier fastball.

What is the test statistic?

Table 10.18 in the book ↗

Pitcher	Sample Mean Speed of Pitches (mph)	Population Standard Deviation
Wesley	86	3
Rodriguez	91	7

Answer

–2.46

10-35 §10.2 source ↗

Use the following information to answer the next five exercises. The mean speeds of fastball pitches from two different baseball pitchers are to be compared. A sample of 14 fastball pitches is measured from each pitcher. The populations have normal distributions. Table 10.18 shows the result. Scouters believe that Rodriguez pitches a speedier fastball.

At the 1 percent significance level, what is your conclusion?

Table 10.18 in the book ↗

Pitcher	Sample Mean Speed of Pitches (mph)	Population Standard Deviation
Wesley	86	3
Rodriguez	91	7

Answer

At the 1 percent significance level, we can reject the null hypothesis. There is sufficient data to conclude that the mean speed of Rodriguez’s fastball is faster than Wesley’s.

10-37 §10.2 source ↗

Use the following information to answer the next five exercises. A researcher is testing the effects of plant food on plant growth. Nine plants have been given the plant food. Another nine plants have not been given the plant food. The heights of the plants are recorded after eight weeks. The populations have normal distributions. The following table is the result. The researcher thinks the food makes the plants grow taller.

State the null and alternative hypotheses.

Table 10.19 in the book ↗

Plant Group	Sample Mean Height of Plants (inches)	Population Standard Deviation
Food	16	2.5
No food	14	1.5

Answer

Subscripts: 1 = Food, 2 = No Food
H₀: μ₁ ≤ μ₂
H_a: μ₁ > μ₂

10-39 §10.2 source ↗

Use the following information to answer the next five exercises. A researcher is testing the effects of plant food on plant growth. Nine plants have been given the plant food. Another nine plants have not been given the plant food. The heights of the plants are recorded after eight weeks. The populations have normal distributions. The following table is the result. The researcher thinks the food makes the plants grow taller.

Draw the graph of the p value.

Table 10.19 in the book ↗

Plant Group	Sample Mean Height of Plants (inches)	Population Standard Deviation
Food	16	2.5
No food	14	1.5

Answer

This is a normal distribution curve with mean equal to zero. The values 0 and 0.1 are labeled on the horiztonal axis. A vertical line extends from 0.1 to the curve. The region under the curve to the right of the line is shaded to represent p-value = 0.0198. — Figure 10.18 in the book ↗

10-41 §10.2 source ↗

Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. Fifteen pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point.

State the null and alternative hypotheses.

Table 10.20 in the book ↗

	Sample Mean Melting Temperatures (°F)	Population Standard Deviation
Alloy Gamma	800	95
Alloy Zeta	900	105

Answer

Subscripts: 1 = Gamma, 2 = Zeta
H₀: μ₁ = μ₂
H_a: μ₁ ≠ μ₂

10-43 §10.2 source ↗

Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. Fifteen pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point.

What is the p value?

Table 10.20 in the book ↗

	Sample Mean Melting Temperatures (°F)	Population Standard Deviation
Alloy Gamma	800	95
Alloy Zeta	900	105

Answer

0.0062

10-45 §10.2 source ↗

Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. Fifteen pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point.

At the 1 percent significance level, what is your conclusion?

Table 10.20 in the book ↗

	Sample Mean Melting Temperatures (°F)	Population Standard Deviation
Alloy Gamma	800	95
Alloy Zeta	900	105

Answer

There is sufficient evidence to reject the null hypothesis. The data support that the melting point for Alloy Zeta is different from the melting point of Alloy Gamma.

10-47 §10.3 source ↗

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.

What is the random variable?

Answer

P′_OS1 – P′_OS2 = difference in the proportions of phones that had system failures within the first eight hours of operation with OS₁ and OS₂.

10-49 §10.3 source ↗

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.

What is the p-value?

Answer

0.1018

10-51 §10.3 source ↗

Use the following information to answer the next 12 exercises. In the recent U.S. Census, 3 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 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.

Is this a test of means or proportions?

Answer

proportions

10-53 §10.3 source ↗

Use the following information to answer the next 12 exercises. In the recent U.S. Census, 3 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 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.

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

Answer

right-tailed

10-55 §10.3 source ↗

Use the following information to answer the next 12 exercises. In the recent U.S. Census, 3 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 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.

In words, define the random variable for this test.

Answer

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

10-57 §10.3 source ↗

Use the following information to answer the next 12 exercises. In the recent U.S. Census, 3 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 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.

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

Answer

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

10-59 §10.3 source ↗

Use the following information to answer the next 12 exercises. In the recent U.S. Census, 3 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 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.

Sketch a graph of the situation. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p-value.

This is a horizontal axis with arrows at each end. The axis is labeled p'N - p'ND — Figure 10.17 in the book ↗

Answer

Check student’s solution.

10-61 §10.3 source ↗

Use the following information to answer the next 12 exercises. In the recent U.S. Census, 3 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 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.

At a preconceived α = 0.05, write the following:

a. Your decision:

b. The reason for your decision:

c. Your conclusion (write out in a complete sentence):

Answer

a. Reject the null hypothesis.

b. p-value < alpha

c. At the 5 percent 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.

10-63 §10.4 source ↗

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 10.21. The before value is matched to an after value, and the differences are calculated. The differences have a normal distribution. Test at the 1 percent significance level.

What is the random variable?

Table 10.21 in the book ↗

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

Answer

the mean difference of the system failures

10-65 §10.4 source ↗

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 10.21. The before value is matched to an after value, and the differences are calculated. The differences have a normal distribution. Test at the 1 percent significance level.

What is the p-value?

Table 10.21 in the book ↗

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

Answer

0.0067

10-67 §10.4 source ↗

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 10.21. The before value is matched to an after value, and the differences are calculated. The differences have a normal distribution. Test at the 1 percent significance level.

What conclusion can you draw about the software patch?

Table 10.21 in the book ↗

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

Answer

With a p-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.

10-69 §10.4 source ↗

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 percent significance level.

What is the p-value?

Table 10.22 in the book ↗

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

Answer

0.0021

10-71 §10.4 source ↗

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 percent significance level.

Draw the graph of the p-value.

Table 10.22 in the book ↗

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

Answer

This is a normal distribution curve with mean equal to zero. The values 0 and 1.67 are labeled on the horiztonal axis. A vertical line extends from 1.67 to the curve. The region under the curve to the right of the line is shaded to represent p-value = 0.0021. — Figure 10.19 in the book ↗

10-73 §10.4 source ↗

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 percent significance level.

State the null and alternative hypotheses.

Table 10.23 in the book ↗

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

Answer

H₀: μ_d ≥ 0

H_a: μ_d < 0

10-75 §10.4 source ↗

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 percent significance level.

What is the p-value?

Table 10.23 in the book ↗

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

Answer

0.0699

10-77 §10.4 source ↗

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 percent significance level.

What is the conclusion?

Table 10.23 in the book ↗

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

Answer

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

10-79 §10.1 source ↗

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 a homework problem in what follows, 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.)

A student at a four-year college claims that mean 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 mean enrollment was 5,068 with a standard deviation of 4,777. Of the 35 four-year colleges surveyed, the mean enrollment was 5,466 with a standard deviation of 8,191.

Answer

Subscripts: 1: two-year colleges, 2: four-year colleges

a. H₀: μ₁ ≥ μ₂

b. H_a: μ₁ < μ₂

c. ${\bar{X}}_{1} - {\bar{X}}_{2}$ is the difference between the mean enrollments of the two-year colleges and the four-year colleges.

d. Student’s t

e. test statistic: -0.2480

f. p-value: 0.4019

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Do not reject.

iii. Reason for Decision: p-value > alpha

iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean enrollment at four-year colleges is higher than at two-year colleges.

10-81 §10.1 source ↗

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 a homework problem in what follows, 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.)

Mean entry-level salaries for college graduates with mechanical engineering degrees and electrical engineering degrees are believed to be approximately the same. A recruiting office thinks that the mean mechanical engineering salary is lower than the mean electrical engineering salary. The recruiting office randomly surveys 50 entry-level mechanical engineers and 60 entry-level electrical engineers. Their mean salaries were $46,100 and $46,700, respectively. Their standard deviations were $3,450 and $4,210, respectively. Conduct a hypothesis test to determine if you agree that the mean entry-level mechanical engineering salary is lower than the mean entry-level electrical engineering salary.

Answer

Subscripts: 1: mechanical engineering, 2: electrical engineering

a. H₀: µ₁ ≥ µ₂

b. H_a: µ₁ < µ₂

c. ${\bar{X}}_{1} - {\bar{X}}_{2}$ is the difference between the mean entry-level salaries of mechanical engineers and electrical engineers.

d. t₁₀₈

e. test statistic: t = –0.82

f. p-value: 0.2061

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Do not reject the null hypothesis.

iii. Reason for Decision: p-value > alpha

iv. Conclusion: At the 5 percent significance level, there is insufficient evidence to conclude that the mean entry-level salaries of mechanical engineers is lower than that of electrical engineers.

10-83 §10.1 source ↗

Use the information from Appendix C to answer the next four exercises.

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

Answer

a. H₀: µ₁ = µ₂

b. H_a: µ₁ ≠ µ₂

c. ${\bar{X}}_{1} - {\bar{X}}_{2}$ is the difference between the mean times for completing a lap in races and in practices.

d. t_20.32

e. test statistic: –4.70

f. p-value: 0.0001

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis.

iii. Reason for Decision: p-value < alpha

iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices.

10-85 §10.1 source ↗

Use the information from Appendix C to answer the next four exercises.

Repeat the test in Exercise 10.83, but this time combine the data from Laps 1 and 5.

Answer

a. H₀: µ₁ = µ₂

b. H_a: µ₁ ≠ µ₂

c. is the difference between the mean times for completing a lap in races and in practices.

d. t_40.94

e. test statistic: –5.08

f. p-value: zero

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis.

iii. Reason for Decision: p-value < alpha

iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices.

10-88 §10.1 source ↗

Use the following information to answer the next two exercises. The Eastern and Western Major League Soccer conferences have a new Reserve Division that allows new players to develop their skills. Data for a randomly picked date showed the following annual goals.

Conduct a hypothesis test to answer the next two exercises.

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

a. There is sufficient evidence to conclude that the W Division teams score fewer goals, on average, than the E teams.

b. There is insufficient evidence to conclude that the W Division teams score more goals, on average, than the E teams.

c. There is insufficient evidence to conclude that the W teams score fewer goals, on average, than the E teams.

d. There is not sufficient evidence to determine a conclusion.

Table 10.25 in the book ↗

Western	Eastern
Los Angeles 9	D 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

Answer

c

10-90 §10.1 source ↗

Use the following information to answer the next two exercises. The Eastern and Western Major League Soccer conferences have a new Reserve Division that allows new players to develop their skills. Data for a randomly picked date showed the following annual goals.

Conduct a hypothesis test to answer the next two exercises.

Elijah wants to know whether textbook costs are different for different courses of study. He selects a random sample of 33 sociology textbooks offered on a popular online site. The mean price of his sample is $74.64 with a standard deviation of $49.36. He then selects a random sample of 33 math and science textbooks from the same site. The mean price of this sample is $111.56 with a standard deviation of $66.90. Is the mean price of a sociology textbook lower than the mean price of a math or science textbook? Test at a 1% significance level.

Table 10.25 in the book ↗

Western	Eastern
Los Angeles 9	D 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

Answer

Test: two independent sample means, population standard deviations unknown.

• µ₁ = the mean price of a sociology text on the selected site

• µ₂ = the mean price of a math/science text on the selected site

Random variable: ${\bar{X}}_{1} - {\bar{X}}_{2}$ = the difference in the sample mean textbook price between sociology texts and math/science texts.

Hypotheses: $H_{0} : µ_{1} - µ_{2} = 0$ , $H_{a} : µ_{1} -$ $µ_{2} <$ $0$ , which can be expressed as $H_{0} : µ_{1} = µ_{2}$ , $H_{a} : µ_{1} < µ_{2}$ .

Distribution for the test: Use t_df, because each sample has more than 30 observations, $d f =$ $n_{1} +$ $n_{2} - 2 =$ $33 +$ $33 - 2 =$ $64$ .

Estimate the critical value on the t-table using the nearest available degrees of freedom, 60. The critical value, 2.660, is found in the .0005 column.

Calculate the test statistic: .

Using a calculator with t_c = –2.55 and df = 64, the left-tailed p-value: 0.0066. Decision: Reject H₀. Conclusion: At the 1% level of significance, from the sample data, there is sufficient evidence to conclude that the mean price of sociology textbooks is less than the mean price of textbooks for math/science.

Distribution: H₀: μ₁ = μ₂, H_a: μ₁ < μ₂
The mean age of entering the industry in Canada is lower than the mean age in the United States.

Graph: left-tailed

p-value : 0.0151

Decision: Do not reject H₀.

Conclusion: At the 1 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of entering the industry in Canada is lower than the mean age in the United States.

This is a normal distribution curve with mean equal to zero. A vertical line near the tail of the curve to the left of zero extends from the axis to the curve. The region under the curve to the left of the line is shaded representing p-value = 0.0157. — Figure 10.20 in the book ↗

10-92 §10.1 source ↗

Table 10.25 in the book ↗

Western	Eastern
Los Angeles 9	D 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

Use the following information to answer the next two exercises. The Eastern and Western Major League Soccer conferences have a new Reserve Division that allows new players to develop their skills. Data for a randomly picked date showed the following annual goals.

Conduct a hypothesis test to answer the next two exercises.

Suppose a statistics instructor believes that there is no significant difference between the mean 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 mean and standard deviation for 35 statistics day students were 75.86 and 16.91, respectively. The mean 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. An appropriate alternative 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

Answer

d

10-94 §10.2 source ↗

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 one of the following 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.)

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 mean 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 you believe that the mean cost for auto insurance for teenage boys is greater than that for teenage girls.

Answer

Subscripts: 1 = boys, 2 = girls

a. H₀: µ₁ ≤ µ₂

b. H_a: µ₁ > µ₂

c. The random variable is the difference in the mean auto insurance costs for boys and girls.

d. normal

e. test statistic: z = 2.50

f. p value: 0.0062

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis.

iii. Reason for Decision: p value < alpha

iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean cost of auto insurance for teenage boys is greater than that for girls.

10-96 §10.2 source ↗

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 one of the following 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.)

Some manufacturers claim that nonhybrid 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 7 mpg. Thirty-one nonhybrid sedans get a mean of 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 evaluate the manufacturers’ claim.

Answer

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

a. H₀: µ₁ ≥ µ₂

b. H_a: µ₁ < µ₂

c. The random variable is the difference in the mean miles per gallon of nonhybrid sedans and hybrid sedans.

d. normal

e. test statistic: 6.36

f. p-value: 0

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis.

iii. Reason for decision: p value < alpha

iv. Conclusion: At the 5 percent 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.

10-98 §10.2 source ↗

Table 10.26 in the book ↗

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

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 one of the following 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.)

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). Table 10.26 contains 10 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).

Answer

a. H₀: µ_d = 0

b. H_a: µ_d < 0

c. The random variable X_d is the average difference between husband’s and wife’s satisfaction level.

d. t₉

e. test statistic: t = –1.86

f. p value: 0.0479

g. Check student’s solution

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis, but run another test.

iii. Reason for Decision: p value < alpha

iv. 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.

10-101 §10.3 source ↗

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 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.)

A 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?

Answer

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

a. H₀: p₁ = p₂

b. H_a: p₁ ≠ p₂

c. The random variable is the difference between the proportions of Hispanic students at Cabrillo College and Lake Tahoe College.

d. normal for two proportions

e. test statistic: 4.29

f. p-value: 0.00002

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis.

iii. Reason for decision: p-value < alpha

iv. Conclusion: There is sufficient evidence to conclude that the proportions of Hispanic students at Cabrillo College and Lake Tahoe College are different.

10-103 §10.3 source ↗

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 percent level of significance, conduct an appropriate hypothesis test.

An appropriate null hypothesis is

a. p₂₀₁₁ ≤ p₂₀₁₀

b. p₂₀₁₁ ≥ p₂₀₁₀

c. μ₂₀₁₁ ≤ μ₂₀₁₀

d. p₂₀₁₁ > p₂₀₁₀

Answer

a

10-105 §10.3 source ↗

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 percent level of significance, conduct an appropriate hypothesis test.

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

Answer

Test: two independent sample proportions.

Random variable: p′₁ - p′₂

Distribution:
H₀: p₁ = p₂
H_a: p₁ ≠ p₂

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

Graph: two-tailed

p-value : 0.0033

Decision: Reject the null hypothesis.

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

This is a normal distribution curve with mean equal to zero. Both the right and left tails of the curve are shaded. Each tail represents 1/2(p-value) = 0.0017. — Figure 10.21 in the book ↗

10-107 §10.3 source ↗

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 percent level of significance, conduct an appropriate hypothesis test.

Two computer users were discussing tablet computers. A higher proportion of people ages 16 to 29 use tablets than of people age 30 and older. Table 10.28 details the number of tablet owners for each age group. Test at the 1 percent level of significance.

Table 10.28 in the book ↗

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

Answer

Test: two independent sample means, population standard deviations unknown.

$μ_{1}$ = the the mean price of a sociology text on the selected site.

$μ_{2}$ = the mean price of a math/science text on the selected site.

Random variable: $\bar{X_{1}} - \bar{X_{1}}$ = the difference in the sample mean textbook price between sociology texts and math/science texts.

Hypotheses: $H_{0} : μ_{1} - μ_{2} =$ $0,$ $H_{a} : μ_{1} - μ_{2} <$ $μ_{2}$ which can be expressed as $H 0 s : μ 1 - μ 2,$ $Ha μ 1 <$ $μ 2$ .

Distribution for the test: Use $t_{d f}$ ; because each sample has more than 30 observations, $d f =$ $n_{1} +$ $n_{2} - 2 =$ $33 +$ $33 - 2 =$ $64$ .

Estimate the critical value on the $t$ -table using the nearest available degrees of freedom, 60. The critical value, 2.660, is found in the .0005 column.

Calculate the test statistic: $t_{c} =$ $\frac{({\bar{X}}_{1} - {\bar{X}}_{2}) - 0}{\sqrt{\frac{{s_{1}}^{2}}{n_{2}} + \frac{{s_{2}}^{2}}{n_{2}}}} =$ $\frac{(74.64 - 111.56) - 0}{\sqrt{\frac{49 . 36^{2}}{33} + \frac{66 . 90^{2}}{33}}} =$ $- 2.55$ .

Using a calculator with $t_{c} = - 2.55$ and $d f = 64$ , the left-tailed $p$ -value: Decision: Reject $H_{0}$ . Conclusion: At the 1% level of significance, from the sample data, there is sufficient evidence to conclude that the mean price of sociology textbooks is less than the mean price of textbooks for math/science.

10-109 §10.3 source ↗

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 percent level of significance, conduct an appropriate hypothesis test.

While her husband spent 2.5 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 do. 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.

Answer

Subscripts: 1: men; 2: women

a. H₀: p₁ ≤ p₂

b. H_a: p₁ > p₂

c. ${P^{'}}_{1} - {P^{'}}_{2}$ is the difference between the proportions of men and women who enjoy shopping for electronic equipment.

d. normal for two proportions

e. test statistic: 0.22

f. p-value: 0.4133

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Do not reject the null hypothesis.

iii. Reason for Decision: p-value > alpha

iv. Conclusion: At the 5 percent 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.

10-111 §10.3 source ↗

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 percent level of significance, conduct an appropriate hypothesis test.

A researcher recently claimed that the proportion of college-age males who wear at least one piece of jewelery is as high as the proportion of college-age females. She conducted a survey in her classes. Out of 107 males, 20 wear at least one piece of jewelery. Out of 92 females, 47 wear at least one piece of jewelery. Do you believe that the proportion of males has reached the proportion of females?

Answer

a. H₀: p₁ = p₂

b. H_a: p₁ ≠ p₂

c. ${P^{'}}_{1} - {P^{'}}_{2}$ is the difference between the proportions of men and women that have at least one pierced ear.

d. normal for two proportions

e. test statistic: –4.82

f. p-value: zero

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis.

iii. Reason for Decision: p-value < alpha

iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the proportions of males and females with at least one pierced ear is different.

10-113 §10.3 source ↗

Table 10.29 in the book ↗

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

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 percent level of significance, conduct an appropriate hypothesis test.

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 10.29, solve our problem.

Answer

a. H₀: µ_d = 0

b. H_a: µ_d > 0

c. The random variable X_d is the mean difference in work times on days when eating breakfast and on days when not eating breakfast.

d. t₉

e. test statistic: 4.8963

f. p-value: 0.0004

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis.

iii. Reason for Decision: p-value < alpha

iv. Conclusion: At the 5 percent 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.

10-114 §10.4 source ↗

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.)

Ten individuals went on a low-fat diet for 12 weeks to lower their cholesterol. The data are recorded in Table 10.30. Do you think that their cholesterol levels were significantly lowered?

Table 10.30 in the book ↗

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

Answer

p-value = 0.1494

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

10-116 §10.4 source ↗

Use the following information to answer the next two exercises. A new preventative medication was tried on a group of 224 patients who had the same risk factors for a disease. 45 patients developed the disease after four years. In a control group of 224 patients, 68 developed the disease after four years. We want to test whether the method of treatment reduces the proportion of patients who develop the disease after four years.

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

If the p-value is 0.0062, what is the conclusion? Use α = 0.05.

a. The method has no effect.

b. There is sufficient evidence to conclude that the method reduces the proportion of patients who develop the disease after four years.

c. There is sufficient evidence to conclude that the method increases the proportion of patients who develop the disease after four years.

d. There is insufficient evidence to conclude that the method reduces the proportion of patients who develop the disease after four years.

Answer

b

10-118 §10.4 source ↗

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.

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

a. 0.0014; There is sufficient evidence to conclude that the blood pressure decreased after the training.

b. 0.0014; There is sufficient evidence to conclude that the blood pressure increased after the training.

c. 0.0155; There is sufficient evidence to conclude that the blood pressure decreased after the training.

d. 0.0155; There is sufficient evidence to conclude that the blood pressure increased after the training.

Answer

c

10-120 §10.4 source ↗

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.

A local research group is studying a chronic disease. They believe the number of cases of the disease is higher in 2013 than in 2012 in the southern United States. The group compared the estimates of new cases by southern state in 2012 and 2013. The results are in Table 10.32.

Table 10.32 in the book ↗

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

Answer

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

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

Distribution: t₁₂

H₀: μ_d = 0 H_a: μ_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-value: 0.0004

Decision: Reject H₀.

Conclusion: At the 5 percent 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.

This is a normal distribution curve with mean equal to zero. A vertical line near the tail of the curve to the right of zero extends from the axis to the curve. The region under the curve to the right of the line is shaded representing p-value = 0.0004. — Figure 10.22 in the book ↗

10-122 §10.4 source ↗

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.

A politician asked his staff to determine whether the underemployment rate in the Northeast decreased from 2011 to 2012. The results are in Table 10.34.

Table 10.34 in the book ↗

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

Answer

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: H₀: μ_d = 0 H_a: μ_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-value: 0.1207

Decision: Do not reject H₀.

Conclusion: At the 5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that there was a decrease in the underemployment rates of the northeastern states from 2011 to 2012.

10-124 §10.BIT source ↗

Use the following information to answer the next 10 exercises. Indicate which of the following choices best identifies the hypothesis test.

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.

Answer

e

10-126 §10.BIT source ↗

Use the following information to answer the next 10 exercises. Indicate which of the following choices best identifies the hypothesis test.

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.

Answer

d

10-128 §10.BIT source ↗

Use the following information to answer the next 10 exercises. Indicate which of the following choices best identifies the hypothesis test.

According to a doctor’s magazine, 75 percent of senior citizens think that yearly checkups are very important. A study is done to verify this.

Answer

f

10-130 §10.BIT source ↗

Use the following information to answer the next 10 exercises. Indicate which of the following choices best identifies the hypothesis test.

A recent survey showed an increase in use of prescription medication among local senior citizens as compared to the national percent. Suppose that a survey of 100 local senior citizens and 100 national senior citizens is conducted to see if the proportion of prescription medication use is higher locally than nationally.

Answer

e

10-132 §10.BIT source ↗

Use the following information to answer the next 10 exercises. Indicate which of the following choices best identifies the hypothesis test.

According to a statistics college professor, 68 percent of his students pass the final exam. A graduate researcher designs a study to determine if this claim is true.

Answer

f

The graduate researcher will be comparing a sample proportion to a population proportion or claim. Thus, the study includes the hypothesis test of a single proportion. A two proportion hypothesis test compares two sample proportions.

10-134 §10.BIT source ↗

Use the following information to answer the next 10 exercises. Indicate which of the following choices best identifies the hypothesis test.

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 after having taken her class. She conducts a hypothesis test. The data are shown in Table 10.37.

This is

a. a test of two independent means.

b. a test of two proportions.

c. a test of a single mean.

d. a test of a single proportion.

Table 10.37 in the book ↗

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

Answer

a

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

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

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