Question 568820: In order to control costs, a company wishes to study the amount of money its sales force pends entertaining clients. The following is a random sample of six entertainment expenses (dinner cots for four people) from expense retorts submitted by members of the sale force.
157 132 109 145 125 139
a. Calculate x/- (this is minus over the x), sē, and s for the expense data. In addition, show that the two different formulas for calculating sē 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 forces.
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)? Explain your answer.
d. Compute and interpret the z-score for each of the sex entertainment expenses.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The following is a random sample of six entertainment expenses (dinner cots for four people) from expense retorts submitted by members of the sale force.
157 132 109 145 125 139
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a. Calculate x/- (this is minus over the x), sē, and s for the expense data. In addition, show that the two different formulas for calculating sē give the same result.
x-bar = 134.5
s = 16.63
s^2 = 276.56
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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 forces.
I'll try one of these. I assume you mean a 68.26% interval around the mean.
That would be 34.13% to the right and to the left of the mean.
Left tail with 0.5-0.3413 = 0.1587 ; z = invNorm(0.1587) = -0.9998
Lowest tolerance limit = -0.9998*16.63+134.5 = 117.87
Highest tolerance limit= +0.9998*16.63+134.5 = 151.13
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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)? Explain your answer.
z(190) = (190-134.5)/16.63 = 3.3373
That's 3 standard deviations above the mean. Yes, worthy of investigation.
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d. Compute and interpret the z-score for each of the six entertainment expenses.
Same procedure as "c".
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cheers,
Stan H.
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