Question

A restaurant wants to test a new in-store marketing scheme in a small number of stores...

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 (5.11, 20.14). Which of the following is the appropriate conclusion? The differences were calculated as (after ad campaign - before ad campaign).

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 coming after the ad campaign.

3) 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.

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

5) 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.

Homework Answers

Answer #1

Given,

The 90% confidence interval to estimate the true average difference in nationwide sales quantity before the ad campaign and after is (5.11, 20.14);

i.e Lower Confidence Limit :LCL = 5.11 and Upper Confidence limit: UCL= 20.14

That means the average difference in sales quantity for all stores is between 5.11 and 20.14

i.e We are 90% confident that the average difference in sales quantity for all stores is positive

The differences were calculated as (after ad campaign - before ad campaign).

i.e

Sales after ad campaign - Sales before ad campaign > 0

Sales after ad campaign > Sales before ad campaign i.e  higher sales coming after the ad campaign  

Therefore , the answer

2) 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.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Question 5: You work for the consumer insights department of a major big box retailer and...
Question 5: You work for the consumer insights department of a major big box retailer and you are investigating the efficacy of a new e-mail marketing campaign. Through the use of e-mail analytics research, you have determined that in a sample of 981 monitored subscribers, 289 of them opened the e-mail within 24 hours of receiving it. What is the 95% confidence interval for the true proportion of all e-mail subscribers that opened the e-mail within 24 hours of receiving...
PART A) The owner of a golf course wants to determine if his golf course is...
PART A) The owner of a golf course wants to determine if his golf course is more difficult than the one his friend owns. He has 25 golfers play a round of 18 holes on his golf course and records their scores. Later that week, he has the same 25 golfers play a round of golf on his friend's course and records their scores again. The average difference in the scores (treated as the scores on his course - the...
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...
The owner of a local golf course wants to estimate the difference between the average ages...
The owner of a local golf course wants to estimate the difference between the average ages of males and females that play on the golf course. He randomly samples 25 men and 28 women that play on his course. He finds the average age of the men to be 36.105 with a standard deviation of 5.507. The average age of the women was 48.973 with a standard deviation of 5.62. He uses this information to calculate a 90% confidence interval...
Question 12 (1 point) The owner of a golf course wants to determine if his golf...
Question 12 (1 point) The owner of a golf course wants to determine if his golf course is more difficult than the one his friend owns. He has 9 golfers play a round of 18 holes on his golf course and records their scores. Later that week, he has the same 9 golfers play a round of golf on his friend's course and records their scores again. The average difference in the scores (treated as the scores on his course...
Automobile manufacturers are interested in the difference in reaction times for drivers reacting to traditional incandescent...
Automobile manufacturers are interested in the difference in reaction times for drivers reacting to traditional incandescent lights and to LED lights. A sample of 19 drivers are told to press a button as soon as they see a light flash in front of them and the reaction time was measured in milliseconds. Each driver was shown each type of light. A 90% confidence interval for the average difference between the two reaction times (traditional - LED) was (-306.28, -50.17). Which...
Researchers in the corporate office of an airline wonder if there is a significant difference between...
Researchers in the corporate office of an airline wonder if there is a significant difference between the cost of a flight on Priceline.com vs. the airline's own website. A random sample of 15 flights were tracked on Priceline and the airline's website and the 90% confidence interval for the mean difference in price (Priceline - Airline Site) was (-64.52, 32.5). Which of the following is the appropriate conclusion? Question 3 options: 1) There is no significant difference between the average...
1) The local swim team is considering offering a new semi-private class aimed at entry-level swimmers,...
1) The local swim team is considering offering a new semi-private class aimed at entry-level swimmers, but needs at minimum number of swimmers to sign up in order to be cost effective. Last year’s data showed that during 8 swim sessions the average number of entry-level swimmers attending was 15. Suppose the instructor wants to conduct a hypothesis test to test the population mean is less than 15, and we know that the variable is normally distributed. The sample size...
A pharmaceutical company is testing a new drug to increase memorization ability. It takes a sample...
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...
A new drug to treat high cholesterol is being tested by pharmaceutical company. The cholesterol levels...
A new drug to treat high cholesterol is being tested by pharmaceutical company. The cholesterol levels for 18 patients were recorded before administering the drug and after. The average difference in cholesterol levels (after - before) was 4.19 mg/dL with a standard deviation of 8.055 mg/dL. Using this information, the calculated 90% confidence paired-t interval is (0.887, 7.493). Which of the following is the best interpretation of this interval? Question 8 options: 1) The proportion of all patients that had...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT