Question

A recent survey of 50 executives who were laid off during a recent recession revealed it took a mean of 26 weeks for them to find another position. The standard deviation of the sample was 6.2 weeks. Construct a 95% confidence interval for the population mean.

Answer 1

Answer: $(24.28,\ 27.72)$

Step-by-step explanation:

Given : Sample size : $n=50$

Sample mean : $\overline{x}=26$

Standard deviation : $\sigma =6.2$

Significance level : $\alpha=1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

Formula to find the confidence interval for population mean :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\=26\pm(1.96)\dfrac{6.2}{\sqrt{50}}\\\\\approx26\pm1.72\\\\=(26-1.72,\ 26+1.72)\\\\=(24.28,\ 27.72)$

Hence, a 95% confidence interval for the population mean = $(24.28,\ 27.72)$