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. Is it reasonable that the population mean is 28 weeks? Justify your answer.

Answer 1

Answer with explanation:

Formula to find the confidence interval for population mean :-

$\overline{x}\pm t^*\dfrac{\sigma}{\sqrt{n}}$

, where $\overline{x}$ = sample mean.

t*= critical z-value

n= sample size.

s= sample standard deviation.

By considering the given question , we have

$\overline{x}= 26$

$s=6.2$

n= 50

Degree of freedom : df = 49  [ df=n-1]

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

Using students's t-distribution table, the critical t-value for 95% confidence =$t_{\alpha/2,df}=t_{0.025,49}=2.010$

Then,  95% confidence interval for the population mean will be :

$26\pm (2.010)\dfrac{6.2}{\sqrt{50}}$

$=26\pm (2.010)\dfrac{6.2}{7.0710}$

$=26\pm (2.010)(0.87682)$

$\approx26\pm1.76$

$=(26-1.76,\ 26+1.76)=(24.24,\ 27.76)$

Hence, a 95% confidence interval for the population mean = (24.24, 27.76)

Since 28 is not contained in the above confidence interval , it means it is not reasonable that the population mean is 28 weeks.