Answer to Question #219979 in Statistics and Probability for Nileman

Question

Assume that you want to construct a 95% confidence interval estimate of a population
mean.Find the estimate of the sample size needed to obtain the specified margin of error
for the 95% confidence interval. The sample standard deviation is given below:
. (Round to the
Margin of error is $7, standard deviation is $24.The required sample size is_
nearest whole number).

Answer 1

The following expression to compute the confidence interval for the mean is used:

$$CI=(\bar{x}-z_c\times\dfrac{\sigma}{\sqrt{n}},\bar{x}+z_c\times\dfrac{\sigma}{\sqrt{n}}),$$

where the critical value correspond to critical values associated to the Normal distribution. The critical values for the given $\alpha=0.05$ is $z_c=z_{1-\alpha/2}=1.96,$

$$margin\ of\ error=E=z_c\times\dfrac{\sigma}{\sqrt{n}}$$

Then

$$n=(\dfrac{z_c\sigma}{E})^2$$

Given $E=7, \sigma=24$

$$n=(\dfrac{1.96(24)}{7})^2=45$$

The required sample size is 45.