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

95% Confidence interval: (24.24 ,27.76)

The population mean 28 weeks is not reasonable.

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(s)/(√(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(2.009)/(√(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.