Question:
Problem 8.20: A recent survey showed that small businesses with annual revenues of less than $ 1,000,000 expend, on the average, approximately $ 158,000 per year on advertising. You are interested in determining whether this figure holds for a small businesses in you region of the country. You conduct a survey of a random sample of 22 companies in your region that have annual revenues of less than $ 1,000,000 and ask company officials to report their annual advertising revenues. The results of the study are a sample average advertising, expenditure of $ 140,000 with a standard deviation of $39,000. Use this sample information to establish a 98% confidence interval for the population mean expenditure. Assume annual advertising expenditures are normally distributed. Now go back and examine the $ 158,000 figure that was reported nationally. Is this figure in your confidence interval? What does that tell you about your region in comparison with the nation?

A.:
Given:
n = 22
mean = 140,000
s = 39,000
98% confidence level = .02%

.02 / 2 = .01%

P = .5000 - .01 = .490

* The 98% level of confidence results in = .01 in each tail of the distribution.
The table t value is: ta/2, n-1 = t.01,21 = 2.518

Solution / Formula:
98% Confidence Interval:

119,063 <= m <= 160,937

98% confident that the mean is in an interval between 119,063 and 160,936.

The $ 158,000 mean average is within the confidence interval. This means that our region is at par with the nation in terms of annual advertising expense.


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