AnswersAQR_A_ICTesting a Claim about a Difference between Proportions

Testing a Claim about a Difference between Proportions — Unit test Answers

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A political pollster claims that 55% of voters prefer candidate A. To investigate this claim, a random sample of 75 voters is polled. The pollster finds that 39 of those polled prefer candidate A. He would like to know if the data provide convincing evidence that the true proportion of all voters who prefer candidate A is less than 55%. What are the values of the test statistic and P-value for this test?Find the z-table here.

A
z = StartStartFraction 0.52 minus 0.55 OverOver StartRoot StartFraction 0.52 (1 minus 0.52) Over 75 EndFraction EndEndFraction, p-value = 0.3015
Option A
B
z = StartStartFraction 0.52 minus 0.55 OverOver StartRoot StartFraction 0.52 (1 minus 0.52) Over 75 EndFraction EndEndFraction, p-value = 0.6030
Option B
C
z = StartStartFraction 0.52 minus 0.55 OverOver StartRoot StartFraction 0.55 (1 minus 0.55) Over 75 EndFraction EndEndFraction, p-value = 0.3015
Option C
D
z = StartStartFraction 0.52 minus 0.55 OverOver StartRoot StartFraction 0.55 (1 minus 0.55) Over 75 EndFraction EndEndFraction, p-value = 0.6030
Option D
2
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The owner of a popular coffee shop wants to determine if there is a difference between the proportion of customers who use their own cups when they purchase a coffee beverage, and the proportion of customers who use their own cups when they purchase an espresso beverage. Customers using their own cups get a 5% discount, which is displayed on the receipt. The owner randomly selects 50 receipts from all coffee purchases and 50 receipts from all espresso purchases. For coffee purchases, 24 receipts showed that the customer used their own cup. For espresso purchases, 18 receipts showed that the customer used their own cup. Assuming the conditions for inference have been met, what is the 99% confidence interval for the difference in proportion of customers who use their own cups?Find the z-table here.

A
(0.48 minus 0.36) plus-or-minus 2.58 StartRoot StartFraction 0.48 (1 minus 0.48) Over 50 EndFraction + StartFraction 0.36 (1 minus 0.36) Over 50 EndFraction EndRoot
Option A
B
(0.48 minus 0.36) plus-or-minus 2.33 StartRoot StartFraction 0.48 (1 minus 0.48) Over 50 EndFraction + StartFraction 0.36 (1 minus 0.36) Over 50 EndFraction EndRoot
Option B
C
(0.52 minus 0.64) plus-or-minus 2.58 StartRoot StartFraction 0.52 (1 minus 0.52) Over 100 EndFraction + StartFraction 0.64 (1 minus 0.64) Over 100 EndFraction EndRoot
Option C
D
(0.52 minus 0.64) plus-or-minus 2.33 StartRoot StartFraction 0.52 (1 minus 0.52) Over 100 EndFraction + StartFraction 0.64 (1 minus 0.64) Over 100 EndFraction EndRoot
Option D
3

A therapist wanted to determine if yoga or meditation is better for relieving stress. The therapist recruited 100 of her high-stress patients. Fifty of them were randomly assigned to take weekly yoga classes, and the other 50 were assigned weekly meditation classes. After one month, 30 of the 50 patients in the yoga group reported less stress, and 35 of the 50 patients in the meditation group reported less stress. Assuming the conditions for inference are met, what is the 95% confidence interval for the difference in proportions of patients experiencing stress relief from the yoga and meditation groups?Find the z-table here.

A
(0.40 minus 0.30) plus-or-minus 1.65 StartRoot StartFraction 0.40 (1 minus 0.40) Over 100 EndFraction + StartFraction 0.30 (1 minus 0.30) Over 100 EndFraction EndRoot
Option A
B
(0.60 minus 0.70) plus-or-minus 1.96 StartRoot StartFraction 0.60 (1 minus 0.60) Over 100 EndFraction + StartFraction 0.70 (1 minus 0.70) Over 100 EndFraction EndRoot
Option B
C
(0.60 minus 0.70) plus-or-minus 1.65 StartRoot StartFraction 0.60 (1 minus 0.60) Over 50 EndFraction + StartFraction 0.70 (1 minus 0.70) Over 50 EndFraction EndRoot
Option C
D
(0.60 minus 0.70) plus-or-minus 1.96 StartRoot StartFraction 0.60 (1 minus 0.60) Over 50 EndFraction + StartFraction 0.70 (1 minus 0.70) Over 50 EndFraction EndRoot
Option D
4

A school guidance counselor is concerned that a greater proportion of high school students are working part-time jobs during the school year than a decade ago. A decade ago, 28% of high school students worked a part-time job during the school year. To investigate whether the proportion is greater today, a random sample of 80 high school students is selected. It is discovered that 37.5% of them work part-time jobs during the school year. The guidance counselor would like to know if the data provide convincing evidence that the true proportion of all high school students who work a part-time job during the school year is greater than 0.28. What are the appropriate hypotheses for this test?

A
H0: p = 0.28 versus Ha: p < 0.28, where p = the proportion of all high school students who work a part-time job during the school year.
B
H0: p = 0.28 versus Ha: p > 0.28, where p = the proportion of all high school students who work a part-time job during the school year.
C
H0: p = 0.28 versus Ha: p < 0.28, where p = the proportion of high school students in the sample who work a part-time job during the school year.
D
H0: p = 0.28 versus Ha: p > 0.28, where p = the proportion of high school students in the sample who work a part-time job during the school year.
8

A school guidance counselor is concerned that a greater proportion of high school students are working part-time jobs during the school year than a decade ago. A decade ago, 28% of high school students worked a part-time job during the school year. To investigate whether the proportion is greater today, a random sample of 80 high school students is selected. It is discovered that 37.5% of them work part-time jobs during the school year. The guidance counselor would like to know if the data provide convincing evidence that the true proportion of all high school students who work a part-time job during the school year is greater than 0.28.

A
If the true proportion of high school students who work a part-time job during the school year is p = 0.747, there is a 0.59 probability that the guidance counselor will find convincing evidence for Ha: p > 0.28.
B
If the true proportion of high school students who work a part-time job during the school year is p = 0.747, there is a 0.59 probability that the guidance counselor will find convincing evidence for Ha: p = 0.28.
C
If the true proportion of high school students who work a part-time job during the school year is p = 0.28, there is a 0.41 probability that the guidance counselor will find convincing evidence for Ha: p > 0.28.
D
If the true proportion of high school students who work a part-time job during the school year is p = 0.28, there is a 0.41 probability that the guidance counselor will find convincing evidence for Ha: p = 0.28.
9

A 95% confidence interval for the true proportion of math students who prefer to use a handheld calculator versus computer software for computations is (0.751, 0.863). Is it reasonable to believe more than 75% of math students prefer to use a handheld calculator versus computer software for computations?

A
Yes, because the entire interval is greater than 0.75.
B
Yes, because the center of the interval is 0.807, which is greater than 0.75.
C
No, because a different sample might give different results.
D
No, because 0.751 is barely greater than 0.75, which is not statistically significant.
10

It is common knowledge that a fair penny will land heads up 50% of the time and tails up 50% of the time. It is very unlikely for a penny to land on its edge when flipped, so a probability of 0 is assigned to this outcome. A curious student suspects that 5 pennies glued together will land on their edge 50% of the time. To investigate this claim, the student securely glues together 5 pennies and flips the penny stack 100 times. Of the 100 flips, the penny stack lands on its edge 46 times. The student would like to know if the data provide convincing evidence that the true proportion of flips for which the penny stack will land on its edge differs from 0.5. The student tests the hypotheses H0: p = 0.50 versus Ha: p ≠ 0.50, where p = the true proportion of all flips for which the penny stack will land on its edge. The conditions for inference are met. The standardized test statistic is z = –0.80 and the P-value is 0.2119. What conclusion should the student make using the α = 0.10 significance level?

A
Because the test statistic is less than α = 0.10, there is convincing evidence that the true proportion of flips for which the penny stack will land on its edge differs from 0.5.
B
Because the P-value is greater than α = 0.10, there is convincing evidence that the true proportion of flips for which the penny stack will land on its edge differs from 0.5.
C
Because the P-value is greater than α = 0.10, there is not convincing evidence that the true proportion of flips for which the penny stack will land on its edge differs from 0.5.
D
Because the test statistic is less than α = 0.10, there is not convincing evidence that the true proportion of flips for which the penny stack will land on its edge differs from 0.5.
12

A computer company wants to determine the proportion of defective computer chips from a day’s production. A quality control specialist takes a random sample of 100 chips from the day’s production and determines that there are 12 defective chips. Assuming all conditions are met, he constructs a 95% confidence interval for the true proportion of defective chips from a day’s production. What are the calculations for this interval?

A
12 plus-or-minus 1.65 StartRoot StartFraction 12 (1 minus 12) Over 100 EndFraction EndRoot
Option A
B
12 plus-or-minus 1.96 StartRoot StartFraction 12 (1 minus 12) Over 100 EndFraction EndRoot
Option B
C
0.12 plus-or-minus 1.65 StartRoot StartFraction 0.12 (1 minus 0.12) Over 100 EndFraction EndRoot
Option C
D
0.12 plus-or-minus 1.96 StartRoot StartFraction 0.12 (1 minus 0.12) Over 100 EndFraction EndRoot
Option D

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