SOLUTION: An auditor of a small business has sampled 50 of 750 accounts. The sample mean total receivables is $525, and the sample standard deviation is $80. Find a 95% confidence interval f
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-> SOLUTION: An auditor of a small business has sampled 50 of 750 accounts. The sample mean total receivables is $525, and the sample standard deviation is $80. Find a 95% confidence interval f
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Question 1097661: An auditor of a small business has sampled 50 of 750 accounts. The sample mean total receivables is $525, and the sample standard deviation is $80. Find a 95% confidence interval for the total sales of the population of account representatives. Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! alpha(a) = 1 - (95/100) = 0.05
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critical probability(p*) = 1 - (a/2) = 0.975
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since our sample size is > 30, we can use the normal distribution and consult the z-tables to find the critical value(CV) associated with p*
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CV is 1.96
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standard error(SE) = standard deviation of the sample / square root(sample size)
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SE = 80 / sqrt(50) = 11.31
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margin of error(ME) = CV * SE
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ME = 1.96 * 11.31 = 22.17
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95% confidence interval = 525 + or - 22.17, which is (502.83, 547.17)
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