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 percent confidence interval for the population mean. (Round your answers to 2 decimal places.)
The confidence interval is between and
Is it reasonable that the population mean is 28 weeks?

Answer 1

Answer:

95% Confidence interval:  (24.24 ,27.76)

The population mean 28 weeks is not reasonable.

Step-by-step explanation:

We are given the following in the question:

Sample mean = 26

Sample size, n = 50

Alpha, α = 0.05

Sample standard deviation, s = 6.2

95% Confidence interval:  

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$  

Putting the values, we get,  

$t_{critical}\text{ at degree of freedom 49 and}~\alpha_{0.05} = \pm 2.009$  

$26 \pm 2.009(\displaystyle\frac{2.009}{\sqrt{50}} ) = 26 \pm 1.76 = (24.24 ,27.76)$

No, it is not reasonable that population mean is 28 weeks because it does not lie in the confidence interval calculated.