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

To calculate the pooled proportion, first calculate the proportion of Fly-More Academy graduates who passed on the first try:

70/140 = 0.5

Then, calculate the proportion of Blue Yonder Institute graduates who passed on the first try:

104/260 = 0.4

To get the pooled proportion, take the weighted average of these two proportions based on their sample sizes:

(140/400) * 0.5 + (260/400) * 0.4 = 0.435

Therefore, the best choice is B, 0.435, as it is the calculated pooled proportion.

Question 2

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 sample sizes are large enough (both are 50), and the number of successes and failures in each group are greater than 5, which justifies the normality assumption for the difference of proportions.

Question 3

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 p-value for a left-tailed test is approximately 0.24. This is because the proportion of arts and sciences seniors who plan to pursue an MBA degree is 0.15 (120/800), which is greater than the proportion of business school seniors who plan to pursue an MBA degree (0.13, or 26/200).

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

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 test statistic for testing the equality of two proportions is given by: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.In this case, p1 = 160/200 = 0.8, p2 = 144/200 = 0.72, n1 = 200, n2 = 200, and p = (160 + 144) / (200 + 200) = 0.76.Plugging these values into the formula, we get:z = (0.8 - 0.72) / sqrt(0.76(1-0.76)(1/200 + 1/200)) = 1.873Therefore, the correct answer is C.

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. 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

The answer of Two well-known aviation training schools are being...

Question 7

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 conducting a right-tailed test to determine if the pass rate at Blue Yonder Institute is higher than that of Fly-More Academy. We can use a two-sample z-test for proportions to compare the two groups.

The observed proportion of Fly-More Academy graduates passing on the first try is 70/140 = 0.5, and the observed proportion of Blue Yonder Institute graduates passing on the first try is 104/260 = 0.4.

The pooled proportion of passing is (70+104)/(140+260) = 0.43.

The test statistic is:
z = (0.4 - 0.5) / sqrt(0.43(1-0.43)(1/140 + 1/260)) = -2.34

Using a standard normal distribution table or calculator, the p-value for this test is approximately 0.027. This means that there is strong evidence to suggest that the pass rate at Blue Yonder Institute is higher than at Fly-More Academy, with a p-value less than the conventional alpha level of 0.05. Therefore, we reject the null hypothesis and conclude that there is a significant difference in the pass rates between the two schools.

Question 8

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 pooled proportion is calculated by combining the successes and total trials from both groups: (160 + 144) / (200 + 200) = 304 / 400 = 0.76.

Question 9

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

The pooled proportion is calculated by taking the total number of successes (in this case, students planning to pursue an MBA) and dividing it by the total number of trials (total students surveyed). Here, $26 + 120 = 146$ students plan to pursue an MBA out of $200 + 800 = 1000$ students surveyed. The pooled proportion is the total successes (146) not divided by the total number of students, so the correct answer is the total number of students planning to pursue an MBA, which is 146.

Question 10

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 11

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 .0275, 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 0.0275 is less than the significance level of 0.025, so we reject the null hypothesis of equal proportions and conclude that the proportion of graduates passing the FAA exam on the first try is significantly different between the two schools. There is no need to change the significance level to 0.05 as it would only increase the chance of a type I error.

Question 12

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 95 percent confidence interval for the difference of proportions would be approximately:
A)[+.013, +.263].
B)[-.014, +.188].
C)[-.003, +.163].
D)[+.057, +.261].

Accepted Answer

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

(point estimate) ± (critical value) × (standard error)

where the point estimate is the difference between the sample proportions (160/200 - 144/200 = 0.08), the critical value is found using a standard normal distribution for a 95% confidence interval (z* = 1.96), and the standard error is given by

sqrt(p1(1-p1)/n1 + p2(1-p2)/n2)

where p1 = proportion of excellent ratings among youths, p2 = proportion of excellent ratings among adults, n1 = sample size of youths, and n2 = sample size of adults.

Plugging in the values, we get a confidence interval of approximately [-0.003, 0.163], which is closest to choice C.

Question 13

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 find the z-test statistic, we first need to calculate the proportions of each group:

Proportion of business school seniors pursuing MBA = 26/200 = 0.13
Proportion of arts and sciences seniors pursuing MBA = 120/800 = 0.15

Next, we need to calculate the standard error of the difference between two proportions:

SE = sqrt(p1(1-p1)/n1 + p2(1-p2)/n2)
SE = sqrt(0.13(0.87)/200 + 0.15(0.85)/800)
SE = 0.0287

Finally, we can calculate the z-test statistic:

z = (p1 - p2) / SE
z = (0.13 - 0.15) / 0.0287
z = -0.716

Therefore, the z-test statistic is -0.716, which corresponds to choice letter C.

Question 14

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

The assumption of normality of the test statistic is clearly justified because the sample sizes for both Fly-More Academy and Blue Yonder Institute are greater than 30, which satisfies the criteria for the normal approximation of the binomial distribution. Therefore, we can use a z-test to compare the two proportions.

Question 15

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 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. 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 when both sample sizes are greater than 10 and the expected number of successes and failures in each group is at least 5. In this case, the sample size for the arts and sciences group is 800, which is clearly greater than 10, and the expected number of successes (120*0.15 = 18) and failures (680*0.85 = 578) in this group are both greater than 5. Therefore, we can assume normality for the difference of proportions and use a z-test to compare the proportions between the two groups.

Question 17

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 critical value for a right-tailed t-test with α = 0.10 and degrees of freedom calculated from the sample sizes and variances is approximately 1.337.

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. $$\begin{array} { | r r l | } 
\hline p 1 & p 2 & \
0.15 & 0.12 & \mathrm { p } \text { (as decimal) } \
120 / 800 & 144 / 1200 & \mathrm { p } \text { (as fraction) } \
120 . & 144 . & \mathrm { X } \
800 & 1200 & \mathrm { n } \
& & \
0.03 & \text { sample difference } & \
0.00 & \text { hypothesized difference } & \
0.01545 & \text { std. error } & \
\text { x.xx } & \text { z } & \
.0522 & \text { p-value (two-tailed) } & \
\hline
\end{array}$$ 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

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. $$\begin{array} { | r r l | } 
\hline p 1 & p 2 & \
0.15 & 0.12 & \mathrm { p } \text { (as decimal) } \
120 / 800 & 144 / 1200 & \mathrm { p } \text { (as fraction) } \
120 . & 144 . & \mathrm { X } \
800 & 1200 & \mathrm { n } \
& & \
0.03 & \text { sample difference } & \
0.00 & \text { hypothesized difference } & \
0.01545 & \text { std. error } & \
\text { x.xx } & \text { z } & \
.0522 & \text { p-value (two-tailed) } & \
\hline
\end{array}$$ At ? = .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 p-value (0.0522) is greater than the significance level (0.05), so we fail to reject the null hypothesis. Therefore, we can conclude that there is not enough evidence to suggest a significant difference in absenteeism rates between the two plants.

Question 20

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).

Quiz 10: Two-Sample Hypothesis Tests

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:

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:

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 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?

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 95 percent confidence interval for the difference of proportions would be approximately:

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?

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: