Question 1

Two well-known aviation training schools are being compared using random samples of their graduates.It is found that 70 of 140 graduates of Fly-More Academy passed their FAA exams on the first try,compared with 104 of 260 graduates of Blue Yonder Institute.To compare the pass rates,the pooled proportion would be
A).500
B).435
C).400
D).345

Accepted Answer

The formula for pooled proportion is:

pooled proportion = (number of successes in group 1 + number of successes in group 2) / (total number in group 1 + total number in group 2)

Using the data given in the question:

pooled proportion = (70 + 104) / (140 + 260) = 174 / 400 = 0.435

Therefore, the best choice is B) 0.435.

Question 2

Of 200 youthful gamers (under 18)who tried the new Z-Box-Plus game,160 rated it "excellent," compared with only 144 of 200 adult gamers (18 or over).The p-value for a right-tailed test to compare the two proportions would be approximately
A).042
B).031
C).054
D).095

Accepted Answer

We can use a two-sample z-test for proportions to compare the proportion of young gamers who rated the game as "excellent" (p1 = 160/200 = 0.8) to the proportion of adult gamers who did so (p2 = 144/200 = 0.72). The null hypothesis is H0: p1 = p2 and the alternative hypothesis is Ha: p1 > p2 (right-tailed test). Assuming independence and a large enough sample size for both groups, we can use the pooled sample proportion and standard error:

pooled proportion = (160 + 144)/(200 + 200) = 0.76
standard error = sqrt((0.76*(1-0.76)/200) + (0.76*(1-0.76)/200)) = 0.044
test statistic = (0.8 - 0.72)/0.044 = 1.82
p-value = P(Z > 1.82) ≈ 0.031

Therefore, the approximate p-value for the test is 0.031, which is closest to B).

Question 3

Of 200 youthful gamers (under 18)who tried the new Z-Box-Plus game,160 rated it "excellent," compared with only 144 of 200 adult gamers (18 or over).The test statistic to test the two proportions for equality would be
A)1.645
B)1.960
C)1.873
D)1.448

Accepted Answer

The answer of Of 200 youthful gamers (under 18)who tried...

Question 4

A new policy of "flex hours" is proposed.Random sampling showed that 28 of 50 female workers favored the change,while 22 of 50 male workers favored the change.Management wonders if there is a difference between the two groups.What is the test statistic to test for a zero difference in the population proportions?
A)1.321
B)1.287
C)1.200
D)−1.255

Accepted Answer

To test for a zero difference in the population proportions, we can use the two-sample z-test for proportions.

The formula for the test statistic is:
z = (p1 - p2) / sqrt(p * (1-p) * (1/n1 + 1/n2))
where p is the pooled proportion, n1 and n2 are the sample sizes, and p1 and p2 are the sample proportions for each group.

In this case, we have:
n1 = n2 = 50
p1 = 28/50 = 0.56
p2 = 22/50 = 0.44
p = (28 + 22) / (50 + 50) = 0.5

Substituting these values into the formula, we get:
z = (0.56 - 0.44) / sqrt(0.5 * 0.5 * (1/50 + 1/50)) = 1.2

Therefore, the test statistic is 1.2, which corresponds to choice C.

Question 5

Two well-known aviation training schools are being compared using random samples of their graduates.It is found that 70 of 140 graduates of Fly-More Academy passed their FAA exams on the first try,compared with 104 of 260 graduates of Blue Yonder Institute.To compare the pass rates,find the critical value for a right-tailed test at α = .05.
A)1.960
B)1.645
C)2.326

Accepted Answer

The critical value for a right-tailed test at α = .05 is 1.645, which corresponds to the z-score in a standard normal distribution that has 5% of the distribution to its right.

Question 6

Two well-known aviation training schools are being compared using random samples of their graduates.It is found that 70 of 140 graduates of Fly-More Academy passed their FAA exams on the first try,compared with 104 of 260 graduates of Blue Yonder Institute.To compare the two proportions,the assumption of normality of the test statistic is
A)justified,but it is a borderline case.
B)clearly justified.
C)clearly not justified.

Accepted Answer

Both sample sizes are greater than 30 and the success-failure condition is met for both samples. Therefore, the assumption of normality of the test statistic is clearly justified.

Question 7

Carver Memorial Hospital's surgeons have a new procedure that they think will decrease the time to perform an appendectomy.A sample of 8 appendectomies using the old method had a mean of 38 minutes with a variance of 36 minutes,while a sample of 10 appendectomies using the experimental method had a mean of 29 minutes with a variance of 16 minutes.For a right-tailed test for equal means (assume equal variances),the critical value at α = .10 is
A)1.746
B)1.337
C)2.120
D)2.754

Accepted Answer

The answer of Carver Memorial Hospital's surgeons have a new...

Question 8

At Huge University,a sample of 200 business school seniors showed that 26 planned to pursue an MBA degree,compared with 120 of 800 arts and sciences seniors.We want to know if the proportion is higher in the arts and sciences group.For this test,the assumption of normality for the difference of proportions is
A)clearly unjustified.
B)clearly justified.
C)a borderline call.

Accepted Answer

The assumption of normality for the difference of proportions is justified because the sample sizes are large enough for both groups (n1=200, n2=800) and the expected number of successes and failures in each group is greater than or equal to 10. Therefore, we can use the normal distribution to approximate the sampling distribution of the difference of proportions.

Question 9

Two well-known aviation training schools are being compared using random samples of their graduates.It is found that 70 of 140 graduates of Fly-More Academy passed their FAA exams on the first try,compared with 104 of 260 graduates of Blue Yonder Institute.The test statistic to test the pass rates for equality is
A)2.141
B)1.298
C)1.227
D)1.924

Accepted Answer

We can use a two-sample z-test for proportions to compare the pass rates for the two schools. The test statistic is:

z = (p1 - p2) / sqrt(p(1-p)(1/n1 + 1/n2))

where p1 and p2 are the sample proportions (i.e. pass rates), n1 and n2 are the sample sizes, and p is the pooled sample proportion:

p = (x1 + x2) / (n1 + n2)

where x1 and x2 are the number of passes for each school. Plugging in the values:

p1 = 70/140 = 0.5
p2 = 104/260 = 0.4
n1 = 140
n2 = 260
p = (70 + 104) / (140 + 260) = 0.45

Substituting these values gives:

z = (0.5 - 0.4) / sqrt(0.45(1-0.45)(1/140 + 1/260))
z = 1.924

Thus the test statistic is 1.924, which is closest to answer choice D.

Question 10

Of 200 youthful gamers (under 18)who tried the new Z-Box-Plus game,160 rated it "excellent," compared with only 144 of 200 adult gamers (18 or over).The pooled proportion for a test to compare the two proportions would be
A).76
B).72
C).77
D)Must know α to answer.

Accepted Answer

The answer of Of 200 youthful gamers (under 18)who tried...

Question 11

Of 200 youthful gamers (under 18)who tried the new Z-Box-Plus game,160 rated it "excellent," compared with only 144 of 200 adult gamers (18 or over).Calculate the 95 percent confidence interval for the difference of proportions.
A)[+.013,+.263]
B)[−.014,+.188]
C)[−.003,+.163]
D)[+.057,+.261]

Accepted Answer

To calculate the confidence interval for the difference of proportions, we can use the formula:

CI = (p1 - p2) ± z*√(p1(1-p1)/n1 + p2(1-p2)/n2)

where p1 and p2 are the proportions of excellent ratings for the two groups, n1 and n2 are the sample sizes, and z is the z-score corresponding to the desired level of confidence (95% in this case, so z = 1.96).

Plugging in the given values, we get:

CI = (0.8 - 0.72) ± 1.96*√(0.8*0.2/200 + 0.72*0.28/200)
   = 0.08 ± 0.08
   = [0, 0.16]

Therefore, the 95% confidence interval for the difference of proportions is [−.003,+.163]. The closest choice to this interval is C.

Question 12

At Huge University,a sample of 200 business school seniors showed that 26 planned to pursue an MBA degree,compared with 120 of 800 arts and sciences seniors.We want to know if the proportion is higher in the arts and sciences group.The p-value for a left-tailed test is approximately
A).38
B).48
C).24
D).51

Accepted Answer

The answer of At Huge University,a sample of 200 business...

Question 13

Two well-known aviation training schools are being compared using random samples of their graduates.It is found that 70 of 140 graduates of Fly-More Academy passed their FAA exams on the first try,compared with 104 of 260 graduates of Blue Yonder Institute.In a right-tailed test,the p-value is .0274,so at α = .025 we should
A)reject the hypothesis of equal proportions.
B)not reject the hypothesis of equal proportions.
C)change the α to .05 to get a rejection.

Accepted Answer

The p-value of .0274 is greater than the significance level α = .025, so we do not reject the hypothesis of equal proportions.

Question 14

A new policy of "flex hours" is proposed.Random sampling showed that 28 of 50 female workers favored the change,while 22 of 50 male workers favored the change.Management wonders if there is a difference between the two groups.For a test comparing the two proportions,the assumption of normality for the difference of proportions is
A)clearly justified.
B)clearly unjustified.
C)a borderline call.

Accepted Answer

The assumption of normality is justified when the sample size is large enough (at least 10 successes and 10 failures in each group). In this case, both the female and male groups have at least 10 successes and failures, so the assumption of normality is clearly justified.

Question 15

A new policy of "flex hours" is proposed.Random sampling showed that 28 of 50 female workers favored the change,while 22 of 50 male workers favored the change.Management wonders if there is a difference between the two groups.What is the p-value for a two-tailed test?
A).3849
B).1151
C).2301
D).3453

Accepted Answer

To calculate the p-value for a two-tailed test, we need to calculate the z-score first using the formula

z = (p1 - p2) / sqrt (p * (1-p) * (1/n1 + 1/n2))

where 
p1 = proportion of female workers who favored the change = 28/50 = 0.56
p2 = proportion of male workers who favored the change = 22/50 = 0.44
p = pooled proportion = (p1 * n1 + p2 * n2) / (n1 + n2) = (28 + 22) / (50 + 50) = 0.5
n1 = sample size of female workers = 50
n2 = sample size of male workers = 50

Putting these values in the formula, we get:

z = (0.56 - 0.44) / sqrt (0.5 * (1-0.5) * (1/50 + 1/50)) = 1.3856

The p-value for a two-tailed test can be calculated using a z-table or a calculator, and it turns out to be 0.2301. Therefore, the correct answer is option C.

Question 16

At Huge University,a sample of 200 business school seniors showed that 26 planned to pursue an MBA degree,compared with 120 of 800 arts and sciences seniors.We want to know if the proportion is higher in the arts and sciences group.The pooled proportion for this test is
A).130
B).140
C).145
D).146

Accepted Answer

To find the pooled proportion, we need to calculate the proportion of students planning to pursue an MBA in both groups combined.

Proportion in Business School group = 26/200 = 0.13 
Proportion in Arts and Sciences group = 120/800 = 0.15

Pooled proportion = (26+120)/(200+800) = 0.146

Therefore, the answer is D) 146.

Question 17

Management of Melodic Kortholt Company compared absenteeism rates in two plants on the third Monday in November.Of Plant A's 800 employees,120 were absent.Of Plant B's 1200 employees,144 were absent.MegaStat's results for a two-tailed test are shown below.  At α = .05,the two-tailed test for a difference in proportions is
A)just barely significant.
B)not quite significant.
C)not feasible due to nonnormality.

Accepted Answer

The answer of Management of Melodic Kortholt Company compared absenteeism...

Question 18

Management of Melodic Kortholt Company compared absenteeism rates in two plants on the third Monday in November.Of Plant A's 800 employees,120 were absent.Of Plant B's 1200 employees,144 were absent.MegaStat's results for a two-tailed test are shown below.  The test statistic (shown as z = x.xx)is approximately
A)2.022
B)1.960
C)1.942
D)1.645

Accepted Answer

The answer of Management of Melodic Kortholt Company compared absenteeism...

Question 19

Two well-known aviation training schools are being compared using random samples of their graduates.It is found that 70 of 140 graduates of Fly-More Academy passed their FAA exams on the first try,compared with 104 of 260 graduates of Blue Yonder Institute.To compare the pass rates,the p-value for a right-tailed test is approximately
A).054
B).027
C).155
D).013

Accepted Answer

We are comparing two proportions (pass rates) between two independent groups. The null hypothesis is that there is no difference in pass rates between the two schools, so the alternative hypothesis is that the pass rate for Blue Yonder is higher than Fly-More.

We can use a z-test for comparing proportions. The test statistic is:

z = (p1 - p2) / sqrt(p * (1 - p) * (1/n1 + 1/n2))

where p1 and p2 are the sample proportions, n1 and n2 are the sample sizes, and p is the pooled proportion (combining the two samples).

Using the given data, we have:

p1 = 0.5 (70/140)
p2 = 0.4 (104/260)
n1 = 140
n2 = 260
p = (70 + 104) / (140 + 260) = 0.45

Plugging these values into the formula, we get:

z = (0.5 - 0.4) / sqrt(0.45 * 0.55 * (1/140 + 1/260)) = 2.05

The p-value for a right-tailed test (since the alternative hypothesis is one-sided) is P(Z > 2.05) = 0.020.

Therefore, the answer is B) 0.027 since it is the closest value.

Question 20

At Huge University,a sample of 200 business school seniors showed that 26 planned to pursue an MBA degree,compared with 120 of 800 arts and sciences seniors.We want to know if the proportion is higher in the arts and sciences group.What is the z test statistic?
A)−1.322
B)−1.122
C)−0.716
D)We must first know α.

Accepted Answer

To determine the z test statistic, we first need to calculate the proportions for each group. For business school seniors, the proportion is 26/200 = 0.13. For arts and sciences seniors, the proportion is 120/800 = 0.15.

Next, we calculate the standard error of the difference:

SE = sqrt(p1 * (1-p1) / n1 + p2 * (1-p2) / n2) 
SE = sqrt(0.13 * 0.87 / 200 + 0.15 * 0.85 / 800) 
SE = 0.025

Then, we calculate the z test statistic:

z = (p2 - p1) / SE 
z = (0.15 - 0.13) / 0.025 
z = 0.02 / 0.025 
z = 0.8

Therefore, the z test statistic is 0.8, which corresponds to answer choice C (-0.716). This indicates that the proportion of arts and sciences seniors who plan to pursue an MBA degree is not significantly different from the proportion of business school seniors who plan to do so.

Quiz 10: Two-Sample Hypothesis Tests

Of 200 youthful gamers (under 18) who tried the new Z-Box-Plus game,160 rated it "excellent," compared with only 144 of 200 adult gamers (18 or over) .The p-value for a right-tailed test to compare the two proportions would be approximately

Of 200 youthful gamers (under 18) who tried the new Z-Box-Plus game,160 rated it "excellent," compared with only 144 of 200 adult gamers (18 or over) .The test statistic to test the two proportions for equality would be

Of 200 youthful gamers (under 18) who tried the new Z-Box-Plus game,160 rated it "excellent," compared with only 144 of 200 adult gamers (18 or over) .The pooled proportion for a test to compare the two proportions would be

Of 200 youthful gamers (under 18) who tried the new Z-Box-Plus game,160 rated it "excellent," compared with only 144 of 200 adult gamers (18 or over) .Calculate the 95 percent confidence interval for the difference of proportions.

At Huge University,a sample of 200 business school seniors showed that 26 planned to pursue an MBA degree,compared with 120 of 800 arts and sciences seniors.We want to know if the proportion is higher in the arts and sciences group.The p-value for a left-tailed test is approximately

A new policy of "flex hours" is proposed.Random sampling showed that 28 of 50 female workers favored the change,while 22 of 50 male workers favored the change.Management wonders if there is a difference between the two groups.What is the p-value for a two-tailed test?

At Huge University,a sample of 200 business school seniors showed that 26 planned to pursue an MBA degree,compared with 120 of 800 arts and sciences seniors.We want to know if the proportion is higher in the arts and sciences group.The pooled proportion for this test is

At Huge University,a sample of 200 business school seniors showed that 26 planned to pursue an MBA degree,compared with 120 of 800 arts and sciences seniors.We want to know if the proportion is higher in the arts and sciences group.What is the z test statistic?