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. Assume the population standard deviation is 6.2 weeks. Does the data provide sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance

Answer 1

Answer:

The pvalue of 0.0113 < 0.05 means that there is sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance

Step-by-step explanation:

The null hypothesis is:

$H_{0} = 28$

The alternate hypotesis is:

$H_{1} < 28$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

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.

This means that $n = 50, X = 26$

Assume the population standard deviation is 6.2 weeks.

This means that $\sigma = 6.2$

Does the data provide sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance

We have to find the pvalue of Z, looking at the z-table, when $\mu = 28$. It if is lower than 0.05, it provides evidence.

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{26 - 28}{\frac{6.2}{\sqrt{50}}}$

$z = -2.28$

$z = -2.28$ has a pvalue of 0.0113 < 0.05.

The pvalue of 0.0113 < 0.05 means that there is sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance