Question

Consider a population with a known standard deviation of 18.6. In order to compute an interval estimate for the population mean, a sample of 58 observations is drawn. [You may find it useful to reference the z table. a. Is the condition that X is normally distributed satisfied? Yes No b. Compute the margin of error at a 99% confidence level. (Round intermediate calculations to at least 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 245 observations. (Round intermediate calculations to at least 4 decimal places. Round "z" value to 3 decimal places and final answer to 2 decimal places.) Margin of error

Answer 1

a)

Since sample size is sufficiently large, that is n ( = 58) > 30 ,

Yes, condition that $\bar{x}$ is normally distributed is satisfied.

b)

Margin of error = Z$\alpha$/2 * $\sigma$ / sqrt(n)

= 2.576 * 18.6 / sqrt(58)

= 6.29

c)

Margin of error = Z$\alpha$/2 * $\sigma$ / sqrt(n)

= 2.576 * 18.6 / sqrt(245)

= 3.06