Consider a population with a known standard deviation of 9.8. In order to comput

Question

Consider a population with a known standard deviation of 9.8. In order to compute an interval estimate for the population mean, a sample of 48 observations is drawn. Use Table 1.

Compute the margin of error at a 95% confidence level. (Round intermediate calculations to 4 decimal places. Round "z-value" to 3 decimal places and final answer to 2 decimal places.)

Compute the margin of error at a 95% confidence level based on a larger sample of 490 observations.(Round intermediate calculations to 4 decimal places. Round "z-value" to 3 decimal places and final answer to 2 decimal places.)

Which of the two margins of error will lead to a wider confidence interval?

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Consider a population with a known standard deviation of 9.8. In order to compute an interval estimate for the population mean, a sample of 48 observations is drawn. Use Table 1.

Explanation / Answer

a) yes

b) std error of mean =std deviation/(n)1/2 =9.8/(48)1/2 =1.4145

for 95% CI ; z =1.96

therefore margin of error =z*std error =2.77

c)

std error of mean =std deviation/(n)1/2 =9.8/(490)1/2 =0.4427

for 95% CI ; z =1.96

therefore margin of error =z*std error =0.87

d)

95% confidence interval with n=48

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