AnswersCA-Statistics and Probability BEstimating the Difference between Two Population Proportions

Estimating the Difference between Two Population Proportions — Unit test Answers

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A teacher would like to estimate the mean number of steps students take during the school day. To do so, she selects a random sample of 50 students and gives each one a pedometer at the beginning of the school day. They wear the pedometers all day and then return them to her at the end of the school day. From this, she computes the 98% confidence interval for the true mean number of steps students take during the school day to be 8,500 to 10,200 steps. Which of these statements is a correct interpretation of the confidence level?

A
Approximately 98% of students take between 8,500 and 10,200 steps per day.
B
The teacher can be 98% confident that the interval from 8,500 to 10,200 captures the true mean number of steps taken by students during the school day.
C
If many random samples of size 50 are selected from all students at this school, approximately 98% of the intervals would capture the true mean number of steps taken during the school day.
D
Approximately 98% of the intervals, based upon random samples of size 50 from all students in the high school, will find that the students at this school take approximately 8,500 to 10,200 steps during the school day.
4

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
5

A political candidate feels that she performed particularly well in the most recent debate against her opponent. Her campaign manager polled a random sample of 400 likely voters before the debate and a random sample of 500 likely voters after the debate. The 95% confidence interval for the true difference (post-debate minus pre-debate) in proportions of likely voters who would vote for this candidate was (–0.014, 0.064). Based on this interval, what conclusion should the candidate make about the proportion of likely voters who would vote for her in the upcoming election?

A
There is a 95% chance that the proportion who would vote for her has increased.
B
The candidate cannot conclude that the proportion of likely voters who would vote for her has increased.
C
It can be stated with 95% confidence that there was no change in the proportion who would vote for her.
D
It can be stated with 95% confidence that there is convincing evidence that the proportion who would vote for her has increased.
7

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
8

A statistics class weighed 20 bags of grapes purchased from the store. The bags are advertised to contain 16 ounces, on average. The class calculated the 90% confidence interval for the true mean weight of bags of grapes from this store to be (15.875, 16.595) ounces. What is the correct interpretation of the 90 percent confidence interval?

A
90% of the bags of grapes in the sample weigh 16 ounces.
B
90% of the bags of grapes in the sample weigh between 15.875 and 16.595 ounces.
C
We are 90% confident that the interval from 15.875 ounces to 16.595 ounces captures the true mean weight of bags of grapes.
D
If many samples of size 20 were taken, and confidence intervals created for each sample, 90% of those confidence intervals would contain the true mean weight of bags of grapes.
10

A major car dealership has several stores in a big city. The owner wants to determine if there is a difference in the proportions of SUVs that are sold at stores A and B. The owner gathers the sales records for each store from the past year. A random sample of 55 receipts from store A shows that 30 of the sales were for SUVs. Another random sample of 60 receipts from store B shows that 45 of the sales were for SUVs.Based on the 99% confidence interval, (–0.43, –0.02), is there convincing evidence of a difference in the proportions of sales that are SUVs for the two stores?

A
There is convincing evidence because the entire interval is below 0.
B
There is convincing evidence because the two sample proportions are different.
C
There is not convincing evidence because another confidence interval that uses a higher level of confidence might contain 0.
D
There is not convincing evidence because two different sample sizes were used. In order to determine a difference, the same number of sale records should be selected from each population.
11

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.

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