Question

1.Time taken for oil change

It is known that the amount of time needed to change the oil on a car is normally distributed with a standard deviation of 5 minutes. The manager of a service shop recorded the amount of time (in minutes) to complete a random sample of 10 oil changes. They are listed below. 11 10 16 15 18 12 25 20 18 24

a. The sample average is: _______ minutes (up to 2 decimal points)

B.The sample standard deviation is: _______ minutes (up to 2 decimal points)

The following information applies to the next two questions:

Compute a 95% interval estimate of the population mean.

1. Lower Confidence Level ___ Minute (up to 2 decimal points)
2. Upper Confidence Level ___ Minutes (up to 2 decimal points)
3. Based on your answer above, if you take your car to this particular shop, your car will be serviced between  and minutes. You are  % certain about this.

4. Suppose that the manager feels that the range of values he obtained above are too wide to attract customers. What can he do to obtain a narrower range of values?

Use a 90% confidence interval

Take a random sample of 100 oil changes

Train his employees well so that the variability in time to change oil reduces

All the above

None of the above

5. A marketing researcher wants to estimate the mean amount spent ($) on Amazon.com by Amazon Prime member shoppers. A random sample of 100 Amazon Prime member shoppers who recently made a purchase on Amazon.com yielded a mean of $1,500 and a standard deviation of $200.

Construct a 90% Confidence Interval estimate for the mean spending for all Amazon Prime shoppers.

What is the Lower Confidence Level $______

What is the Upper Confidence Level $______

8. Based on the above calculation, which one of the following statements is correct

We are 90% confident that an Amazon Prime Member spends $1500

We are 90% confident that an Amazon Prime Member spends between $1467.10 and $1532.90

Both of the above statements are true

An Amazon Prime member spends between $1467.10 and $1532.90

9. The researcher is not happy with the estimate, and she wants tighter interval, i.e., a smaller level of error. If she wants the estimate to be within ±$25 with 90% confidence, what sample size does she need?

Sample size _____ (report the next whole number, 100.2 should be reported as 101)

10. The salaries of graduates from the MBA program of a Big Ten school are NOT normally distributed. In order to get a better understanding of the range of salaries made by the graduates, the marketing director compiles a five-year record of salaries offered to students at campus recruitment events. He randomly selects groups of 120 students graduating in Fall, Summer, and Spring for the past five years (i.e., 15 groups, each with 120 students).

According to the Central Limit Theorem, the salaries within any of these 15 groups will be distributed normally. True or False?

According to the Central Limit Theorem, the average salaries of these 15 groups will be distributed normally. True or False

Answer 1

1) a) The sample average () = (11 + 10 + 16 + 15 + 18 + 12 + 25 + 20 + 18 + 24)/10 = 16.9

b) The sample standard deviation(s) = sqrt(((11 - 16.9)^2 + (10 - 16.9)^2 + (16 - 16.9)^2 + (15 - 16.9)^2 + (18 - 16.9)^2 + (12 - 16.9)^2 + (25 - 16.9)^2 + (20 - 16.9)^2 + (18 - 16.9)^2 + (24 - 16.9)^2)/9) = 5.15

At 95% confidence interval the critical value is z_0.025 = 1.96

The 95% confidence interval is

+/- z_0.025 *

= 16.9 +/- 1.96 * 5/

= 16.9 +/- 3.099

= 13.801, 19.999

= 13.80, 20.00

Lower confidence level = 13.80

Upper confidence level = 20.00

If we take our car to this particular shop , our car will be served between 13.80 and 20.00 minutes. We are 95% certain about this.

Option - C) All the above

We can narrower the range by decreasing the confidence level or by increasing the sample size or by decreasing the standard deviation.

At 90% confidence interval the critical value is z_0.05 = 1.645

The 90% confidence interval is

+/- z_0.05 *

= 1500 +/- 1.645 * 200/

= 1500 +/- 32.9

= 1467.10, 1532.90

The lower confidence level = 1467.10

The upper confidence level is 1532.90

8) We are 90% confident that an Amazon prime member spends between $1467.10 and $1532.90.

Margin of error = 25

or, z_0.05 * = 25

or, 1.645 * 200/ = 25

or, n = (1.645 * 200/25)^2

or, n = 174

a) True

Because here the sample size will be 120. According to the central limit theorem since the sample size is greater than 30, so the sampling distribution of the sample mean will be normally distributed.

b) False

Because here the sample size will be 15. Since the sample size is less than 30, so the sampling distribution of the sample mean will not be normally distributed.