Question 13

Suppose you work for Fender Guitar company and you are responsible for testing the integrity of a new formulation of guitar strings. To perform your analysis, you randomly select 46 'high E' strings and put them into a machine that simulates string plucking thousands of times per minute. Sami record the number of plucks each string takes before failure and compile a dataset. Sami find that the average number of plucks is 5,924.7 with a standard deviation of 162.48. A 95% confidence interval for the average number of plucks to failure is (5,876.45, 5,972.95). From the option listed below, what is the appropriate interpretation of this interval?

Question 13 options:

1)We are 95% confident that the average number of plucks to failure for all 'high E' strings tested is between 5,876.45 and 5,972.95.

2)We cannot determine the proper interpretation of this interval.

3)We are 95% confident that the proportion of all 'high E' guitar strings fail with a rate between 5,876.45 and 5,972.95.

4)We are certain that 95% of the average number of plucks to failure for all 'high E' strings will be between 5,876.45 and 5,972.95.

5)We are 95% confident that the average number of plucks to failure for all 'high E' strings is between 5,876.45 and 5,972.95.

Question 14

You read an article in Golf Digest that someone with your average drive distance will have an average score of 83.5. Zack keep track of his scores for the next 13 rounds and see that your average score is 82.85 with a standard deviation of 1.621 strokes. Zack should make 95% confidence interval for your average score and it is (81.87, 83.83). Given this information, what is the best conclusion?

Question 14 options:

1)The average score does not significantly differ from 83.5.

2)We are 95% confident that your average score is greater than 83.5.

3)The percentage of rounds in which you score more than 83.5 is 95%.

4)We cannot determine the proper interpretation based on the information given.

5)We are 95% confident that your average score is less than 83.5.

Question 15

A local pizza place claims that they average a delivery time of 8.16 minutes. To test this claim, you order 12 pizzas over the next month at random times on random days of the week. Sara calculate that the average delivery time is 8.63 minutes with a standard deviation of 1.627 minutes. Sara create a 95% confidence interval of (7.596, 9.664). Of those listed below, what is the best conclusion she can make?

Question 15 options:

1)We cannot determine the proper interpretation based on the information given.

2)The average delivery time does not significalty differ from 8.16 minutes.

3)The percentage of pizzas that arrive around 8.16 minutes is 95%.

4)You are 95% confident that the average delivery time is greater than 8.16 minutes.

5)You are 95% confident that the average delivery time is less than 8.16 minutes.

Question 16

A professor at a university wants to estimate the average number of hours of sleep students get during exam week. The professor wants to find a sample mean that is within 1.965 hours of the true average for all college students at the university with 99% confidence. If the professor knows the standard deviation of the sleep times for all college students is 5.512, what sample size will need to be taken?

Question 16 options:

1)53

2)We do not have enough information to answer this question since we were not given the sample mean.

3)52

4)63

5)58

Question 17

As an avid golfer, a player wants to estimate your average score for 18 holes of golf. Suppose he knows that the standard deviation of his score is 26.226 strokes and he want to find a sample mean that is within 7.534 strokes of his true average for all rounds of golf with 95% confidence. How many rounds would he need to play to determine this?

Question 17 options:

1)57

2)52

3)46

4)47

5)We do not have enough information to answer this question since we were not given the sample mean.

Question 18

The owner of a local supermarket wants to estimate the difference between the average number of gallons of milk sold per day on weekdays and weekends. The owner samples 21 weekdays and finds an average of 279.944 gallons of milk sold on those days with a standard deviation of 34.591. 24 Saturdays and Sundays are sampled and the average number of gallons sold is 367.23 with a standard deviation of 46.774. If a 95% confidence interval is calculated to estimate the difference between the average number of gallons sold on weekdays and weekends, what is the margin of error? Assume both population standard deviations are equal.

Question 18 options:

1)20.874

2)12.417

3)2.017

4)24.337

5)25.045

Question 19

Nansi own a small storefront retail business and are interested in determining the average amount of money a typical customer spends per visit to Nansi store. Nansi takes a random sample over the course of a month for 13 customers and find that the average dollar amount she spent per transaction per customer is $81.702 with a standard deviation of $11.0528. Create a 90% confidence interval for the true average spent for all customers per transaction.

Question 19 options:

1)(-76.2384, 87.1656)

2)(76.2384, 87.1656)

3)(79.9197, 83.4843)

4)(78.6365, 84.7675)

5)(76.2729995, 87.1310005)

Question 20

A pharmaceutical company is testing a new drug to increase memorization ability. It takes a sample of individuals and splits them randomly into two groups. After the drug regimen is completed, all members of the study are given a test for memorization ability with higher scores representing a better ability to memorize. Those 20 participants on the drug had an average test score of 33.489 (SD = 4.319), while those 27 participants not on the drug had an average score of 34.557 (SD = 4.418). You use this information to create a 90% confidence interval for the difference in average test score of (-3.236, 1.1). Which of the following is the best interpretation of this interval?

Question 20 options:

1)We do not know the population means so we do not have enough information to make an interpretation.

2)We are certain that the difference between the average test score of all people who would take drug and all people not taking the drug is between -3.236 and 1.1.

3)We are 90% confident that the difference between the average test score of all people who would take drug and all people not taking the drug is between -3.236 and 1.1.

4)We are 90% sure that the average difference in test score of all people who would take the drug and people not on the drug in the state the study was completed in is between -3.236 and 1.1.

5)We are 90% confident that the difference between the average test score of people who would take drug in the study and people not on the drug in the study is between -3.236 and 1.1.

Question 21

A new drug to treat high cholesterol is being tested by pharmaceutical company. The cholesterol levels for 21 patients were recorded before administering the drug and after. The mean difference in total cholesterol levels (after - before) was 1.053 mg/dL with a standard deviation of 5.722 mg/dL. When creating a 90% confidence interval for the true average difference in cholesterol levels by the drug, what is the margin of error?

Question 21 options:

1)1.6522

2)1.2486

3)2.1536

4)0.6004

5)2.1486

Question 22

A restaurant wants to test a new in-store marketing scheme in a small number of stores before rolling it out nationwide. The new ad promotes a premium drink that they want to increase the sales of. 8 locations are chosen at random and the number of drinks sold are recorded for 2 months before the new ad campaign and 2 months after. The average difference in the sales quantity (after - before) is -87.7 with a standard deviation of 56.25. Calculate a 99% confidence interval to estimate the true average difference in nationwide sales quantity before the ad campaign and after.

Question 22 options:

1)(-107.59, -67.81)

2)(-154.43, -20.97)

3)(-157.3, -18.1)

4)(-91.2, -84.2)

5)(157.3, -18.1)

Question 23

A new drug to treat high cholesterol is being tested by pharmaceutical company. The cholesterol levels for 25 patients were recorded before administering the drug and after. The average difference in cholesterol levels (after - before) was 3.59 mg/dL with a standard deviation of 7.101 mg/dL. Using this information, the calculated 90% confidence paired-t interval is (1.16, 6.02) Which of the following is the best interpretation of this interval?

Question 23 options:

1)The proportion of all patients that had a difference in cholesterol levels between those on the drug and those who are not is 90%.

2)We are 90% confident that the difference between the average cholesterol level for those on the drug and the average cholesterol level for those not on the drug is between 1.16 and 6.02.

3)We are certain the the average difference in cholesterol levels between those who would take the drug and those who would not is between 1.16 and 6.02.

4)We are 90% confident that the average difference in the cholesterol levels of the patients sampled is between 1.16 and 6.02.

5)We are 90% confident that the average difference in cholesterol levels between those who would take the drug and those who would not is between 1.16 and 6.02.

Question 24

A restaurant wants to test a new in-store marketing scheme in a small number of stores before rolling it out nationwide. The new ad promotes a premium drink that they want to increase the sales of. 18 locations are chosen at random and the number of drinks sold are recorded for 2 months before the new ad campaign and 2 months after. The 90% confidence interval to estimate the true average difference in nationwide sales quantity before the ad campaign and after is (-27.87, -10.7). Which of the following is the appropriate conclusion? The differences were calculated as (after ad campaign- before ad campaign).

Question 24 options:

1)We are 90% confident that the average difference in sales quantity for all stores is negative, with the higher sales coming after the ad campaign.

2)We are 90% confident that the average difference in sales quantity for all stores is positive, with the higher sales being before the ad campaign.

3)There is not a significant difference between the average sales quantity before or after the ad campaign.

4)We are 90% confident that the average difference in sales quantity for all stores is negative, with the higher sales being before the ad campaign.

5)We are 90% confident that the average difference in sales quantity for all stores is positive, with the higher sales coming after the ad campaign.