Consider a population with a known standard deviation of 13.6. In order to compu

Question

Consider a population with a known standard deviation of 13.6. In order to compute an interval estimate for the population mean, a sample of 40 observations is drawn. Use Table 1. a. Is the condition that 2formula80.mml is normally distributed satisfied? Yes No b. Compute the margin of error at a 99% confidence level. (Round intermediate calculations to 4 decimal places. Round "z-value" to 3 decimal places and final answer to 2 decimal places.) Margin of error c. Compute the margin of error at a 99% confidence level based on a larger sample of 240 observations. (Round intermediate calculations to 4 decimal places. Round "z-value" to 3 decimal places and final answer to 2 decimal places.) Margin of error d. Which of the two margins of error will lead to a wider confidence interval? 99% confidence with n = 240. 99% confidence with n = 40.

Explanation / Answer

a) here as sample size is greater then 30 therefore condition that sample is normally distributed satisfied

b)std errror =std deviation/(n)1/2 =13.6/(40)1/2 =2.1503

for 99% CI ; z =2.576

hence margin of errror =z*Std error =5.54

c)

std errror =std deviation/(n)1/2 =13.6/(240)1/2 =0.8779

for 99% CI ; z =2.576

hence margin of errror =z*Std error =2.26

d) 99% confidence with n = 40 has larger margin of error

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