Question

This is a multi-part question.

An auditing company published a recent report on restaurant bills over a 12-month period. The independent review indicated that, on average, errors in restaurant bills resulted in an overpayment of $11.45 per day. Concerned about overpayments occurring at his small chain of restaurants, the owner of Prominent Restaurant & Bar asked his general manager to conduct a survey of the bills. The restaurant manager’s survey yielded the following data:

7.45	8.03	8.19	8.22	8.59	8.70
10.29	10.53	10.56	10.61	10.78	10.88
8.82	9.29	9.34	9.55	9.76	9.87
11.01	11.30	11.53	11.68	12.36	12.39
9.99	10.21	10.23
12.40	12.52	13.25

a) Determine the sample mean, the standard deviation as a measure of variation, prepare a histograph (LCL=$7.50, UCL, width = $1, 5 classes maximum).

b) Determine the 5-number summary, and construct a modified box plot.

c) Using the summary statistics and graphs prepared, provide a short paragraph about the data (consider using CVDOT as a guide). Is it normally distributed or not? What is the distribution of the sample data given?

d) Given the sample data, construct a 93% confidence interval for the average overpayment.

e) Conduct a hypothesis test with ? = 0.05 to determine if the average (mean) overpayment is smaller than that indicated by the independent review.

f) Given the results from the hypothesis test in [f.], what conclusion would the restaurant manager present to the owners of Prominent Restaurant & Bar?

g) Consider that a revised report stated that instead of $11.45, the true average overpayment is $10 with an actual standard deviation of $1.75.   What is the probability that a sample of 30 restaurant bills will have an average overpayment less than the sample mean from this sample? Should the restaurant manager and the owner be concerned about overpayment bills?

Answer 1

7.45	8.03	8.19	8.22	8.59	8.7
10.29	10.53	10.56	10.61	10.78	10.88
8.82	9.29	9.34	9.55	9.76	9.87
11.01	11.3	11.53	11.68	12.36	12.39
9.99	10.21	10.23
12.4	12.52	13.25
MEAN	10.2777
S.D	1.4738
MIN	7.4500
MAX	13.2500
Q1	9.1725
MEDIAN	10.2600
Q3	11.3575
RANGE	5.8000

C . I	f
7.50 - 8.50	3
8.50 - 9.50	5
9.50 - 10.50	7
10.50 - 11.50	7
11.50 - 12.50	5
TOTAL	27

NOTE: Actually we have 30 number values. In that the minmum value is 7.45 and maximum value is 13.25; but as per the demand of question here we constructed the frequency distribution with least lowe limit as 7.5 and with the maximum number of 5 class intervals. So we got total as 27 not 30.

(b) FIVE NUMBER SUMMARY:

MIN	7.4500
MAX	13.2500
Q1	9.1725
MEDIAN	10.2600
Q3	11.3575

(C) With the basis of Both charts; we can conlcude that the given data is Normally Distributed.

Especially if you observe the Histogram. the bars gradually increased and reached a Peak Point; from there its gradually decreased with equal proportion of the left side the area.

(d) The 93% confidence interval of is

s = S.D = 1.4738

n = 30

Therefore

Given confidence interval = 93%; So

Therefore

Therefore The 93% confidence interval of is