A survey of 100 businesses revealed that the mean after-tax prot was $80,000 and

Question

A survey of 100 businesses revealed that the mean after-tax prot was $80,000 and the standard deviation was $12,000. You would like to formally estimate the population mean after-tax prot of a business. Q16 Suppose that you use the sample mean as the estimator. Find the standard error of the estimator. (a) 80,000 (b) 12,000 (c) 1,200 (d) 120 (e) 800 Answer (c) 1200 Q17 Find the 95% condence interval for the population mean. (a) [76904, 83096] (b) [77204, 82796] (c) [77648, 82352] (d) [78032, 81968] (e) [78464, 81536] Answer (c) Q18 You want to improve precsion of the interval estimator by making the margin of error as low as 1,500 at the condence level of 99%. What is the minimum required sample size for this level of precision? (a) 427 (b) 347 (c) 246 (d) 173 (e) 105 Answer (a) 8

Explanation / Answer

16) standard error = SD/sqrt(n)

= 12000/sqrt (100)

= 12000/10 = 1200

Option-C is correct.

17) At 95% confidence interval the critical value is Z0.975 = 1.96

Cinfidence interval is

Mean +/- Z0.975 * SD/sqrt (n)

= 80000 +/- 1.96 * 12000/sqrt(100)

= 80000 +/- 2352

= 77648, 82352

Option-C is the correct answer

18) At 99% cinfidence interval the critical value is Z0.995 = 2.58

Margin of error = 1500

Or, Z0.995 * SD/sqrt (n ) = 1500

Or, 2.58 * 12000/sqrt(n) = 1500

Or, sqrt (n) = (2.58 * 12000)/1500

Or, sqrt (n) = 20.64

Or, n = 426.0096 = 426

Option-A is the correct answer.

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