Question

# Question 1 (1 point) You own a small storefront retail business and are interested in determining...

Question 1 (1 point)

You own a small storefront retail business and are interested in determining the average amount of money a typical customer spends per visit to your store. You take a random sample over the course of a month for 8 customers and find that the average dollar amount spent per transaction per customer is \$106.745 with a standard deviation of \$13.7164. Create a 95% confidence interval for the true average spent for all customers per transaction.

Question 1 options:

 1) ( 95.278 , 118.212 )
 2) ( 101.896 , 111.594 )
 3) ( -95.278 , 118.212 )
 4) ( 104.38 , 109.11 )
 5) ( 95.562 , 117.928 )

Question 2 (1 point)

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 59 'high E' strings and put them into a machine that simulates string plucking thousands of times per minute. You record the number of plucks each string takes before failure and compile a dataset. You find that the average number of plucks is 6,810.9 with a standard deviation of 152.04. A 90% confidence interval for the average number of plucks to failure is (6,777.8, 6,844). From the option listed below, what is the appropriate interpretation of this interval?

Question 2 options:

 1) We are 90% confident that the average number of plucks to failure for all 'high E' strings is between 6,777.8 and 6,844.
 2) We are certain that 90% of the average number of plucks to failure for all 'high E' strings will be between 6,777.8 and 6,844.
 3) We cannot determine the proper interpretation of this interval.
 4) We are 90% confident that the average number of plucks to failure for all 'high E' strings tested is between 6,777.8 and 6,844.
 5) We are 90% confident that the proportion of all 'high E' guitar strings fail with a rate between 6,777.8 and 6,844.

Question 3 (1 point)

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. One group is administered the drug, and the other is given a placebo. 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. You are presented a 90% confidence interval for the difference in population mean scores (with drug - without drug) of (-8.63, 6.33). What can you conclude from this interval?

Question 3 options:

 1) We are 90% confident that the average memorization ability of those not on the drug is higher than those who are on the drug.
 2) We do not have enough information to make a conclusion.
 3) There is no significant difference between the average memorization abilities for those on the drug compared to those not on the drug.
 4) We are 90% confident that the difference between the two sample means falls within the interval.
 5) We are 90% confident that the average memorization ability of those on the drug is higher than those who are not on the drug.

Question 4 (1 point)

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 average difference in nationwide sales quantity before the ad campaign to after (after - before) is 3.7 with a standard deviation of 9.01. Using this information, they calculate a 90% confidence paired-t interval of (0.01, 7.39). Which of the following is the best interpretation?

Question 4 options:

 1) We are 90% confident that the difference between the average sales after the ad campaign and the average sales before the ad campaign is between 0.01 and 7.39.
 2) We are certain the average difference in sales quantity between after the ad campaign to before for all stores is between 0.01 and 7.39.
 3) The proportion of all stores that had a difference in sales between after the ad campaign to before is 90%.
 4) We are 90% confident that the average difference in the sales quantity after to before of the stores sampled is between 0.01 and 7.39.
 5) We are 90% confident that the average difference in sales quantity between after the ad campaign to before for all restaurants is between 0.01 and 7.39.

1)

 sample mean 'x̄= 106.745 sample size   n= 8 sample std deviation s= 13.716 std error 'sx=s/√n= 4.8495
 for 95% CI; and 7 df, value of t= 2.365 margin of error E=t*std error    = 11.469 lower bound=sample mean-E = 95.28 Upper bound=sample mean+E = 118.21 from above 95% confidence interval for population mean =(95.28,118.21)

option 1 is correct

2)

We are 90% confident that the average number of plucks to failure for all 'high E' strings is between 6,777.8 and 6,844.

3)

 There is no significant difference between the average memorization abilities for those on the drug compared to those not on the drug.

4)

We are 90% confident that the average difference in sales quantity between after the ad campaign to before for all restaurants is between 0.01 and 7.39.

#### Earn Coins

Coins can be redeemed for fabulous gifts.