Question

In order to control costs, a company wishes to study the amount of money its sales force spends entertaining clients. The following is a random sample of six entertainment expenses (dinner costs for four people) from expense reports submitted by members of the sales force. $157 $132 $109 $145 $125 $139 a. Calculate x , s2, and s for the expense data. In addition, show that the two different formulas for calculating s2 give the same result. b. Assuming that the distribution of entertainment expenses is approximately normally distributed, calculate estimates of tolerance intervals containing 68.26 percent, 95.44 percent, and 99.73 percent of all entertainment expenses by the sales force. c. If a member of the sales force submits an entertainment expense (dinner cost for four) of $190, should this expense be considered unusually high (and possibly worthy of investigation by the company)

Answer 1

Answer:

A) x¯ = 134.5

s² = 276.7 is the same for both formulas

s = 16.634

B) At 68.26%, we have (117.866, 151.134)

At 95.44%, we have (101.232, 167.768)

At 99.73%, we have (84.598, 184.402)

C) Yes, it will be considered high because 190 doesn't fall between any of the 3 tolerance intervals gotten in B above.

Step-by-step explanation:

A) The random 6 entertainment expenses are;

$157, $132, $109, $145, $125, $139

We want to find x¯ which is the mean.

Thus;

x¯ = (Σx)/n

x¯ = (157 + 132 + 109 + 145 + 125 + 139)/6

x¯ = 134.5

s² is the variance and can be calculated from 2 different formulas which are;

s² = (Σ(x - x¯)²)/n

Or s² = ((Σx²) - (Σx)²/n)/(n - 1)

For the first formula, when calculated we have;

s² = [(157−134.5)² + (132−134.5)² +(109−134.5)² + (145−134.5)² +(125−134.5)² + (139−134.5)²]/6

s² = 276.7

For the second formula;

(Σx²) = (157)² + (132)² + (109)² + (145)² +(125)² + (139)² = 109925

(Σx) = (157 + 132 + 109 + 145 + 125 + 139) = 807

Thus;

s² = (109925 - (807²/6))/(6 - 1)

s² = 276.7

So both formulas gave the same result of s²

s = √s² = √276.7

s = 16.634

B) from the empirical rule, 68% of the data will fall within one standard deviation while 95% of the data will fall within two standard deviations, and then 99.7% of the data will fall within three standard deviations from the mean.

Now, 68.26% will correspond to 1 standard deviation from the means and the tolerance level Formula here is;

x¯ ± s = 134.5 ± 16.634

This gives; (117.866, 151.134)

95.44% will correspond to within 2 standard deviations and the formula here is;

x¯ ± 2s = 134.5 ± 2(16.634)

>> (101.232, 167.768)

99.73% will correspond to within 3 standard deviations and the formula here is;

x¯ ± 3s = 134.5 ± 3(16.634)

>> (84.598, 184.402)

C) Yes it would be considered high because it doesn't fall between any of the 3 intervals in B above.