Question

Women at Work. In the article "Job Mobility and Wage Growth" (Monthly Labor Review. Vol. 128. No. 2, pp. 33-39).

A. Light examined data on employment and answered questions regarding why workers separate from their employers. According to the article, the standard deviation of the length of time that women with one job are employed during the first 8 years of their career is 92 weeks. Length of time employed during the first 8 years of a career is a left-skewed variable. For that variable, do the following tasks.

a. Determine the sampling distribution of the sample mean for simple random samples of 50 women with one job. Explain your reasoning.

b. Obtain the probability that the sampling error made in estimating the mean length of time employed by all women with one job by that of a random sample of 50 such women will be at most 20 weeks.

Short answer

Part (a) For simple random samples of $50$women, the sampling distribution of the sample mean is $13.01days$

Part (b) The sample mean of samples of size $50$ has a probability of sampling error of at most $20$ weeks in $0.8757556$ the population mean length of time employed by all women with one job.

Step-by-step solution

Part (a) Step 1: Given information

The standard deviation of the lengths of time women with one job worked over the last eight years is 92 weeks.

population S.D $\sigma =92$week

Let the population mean is $\mu$ weeks

Part (a) Step 2: Concept

$\text{population mean and standard deviation:}{\mu }_{\overline{\tilde{x}}}=\mu \text{and}{\sigma }_{\overline{\tilde{x}}}=\sigma /\sqrt{n}\text{.}$

Part (a) Step 3:Calculation

Sample size $n=50$

We can consider the sample size to be large because it is larger than $30$

As a result of using the C1.T sample, the mean $\overline{x}$follows a normal distribution with a mean$\mu$and S.D.

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{92}{\sqrt{50}}\text{Days} \\ =13.01\text{Days} \end{gathered}$$

Part (b) Step 1: Calculation

We have to find $P(\mu -20\le \overline{X}\le \mu +20)$

Where $\overline{X}~N\left(\mu ,{\sigma }_{\overline{X}}^2\right)$

Where ${\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{n}}=13.01$days

$$\begin{gathered} P[\mu -20\le \overline{X}\le \mu +20] \\ =P\left[\frac{\mu -20-\mu }{{\sigma }_{\overline{X}}}\le \frac{\overline{X}-\mu }{{\sigma }_{\overline{X}}}\le \frac{\mu +20-\mu }{{\sigma }_{\overline{X}}}\right] \\ =P\left[\frac{-20}{13.01}\le z\le \frac{20}{13.61}\right],z=\frac{\overline{X}-\mu }{{\sigma }_{\overline{X}}}~N(0,1) \\ =P[-1.5372\le z\le 1.5372] \\ =\Phi (1.5372)-\Phi (-1.5372) \\ =2\Phi (1.5372)-1\left[\Phi \right(-x)=1-\Phi (x\left)\right] \\ =0.8757556 \end{gathered}$$

As a result, the sample mean of samples of size $50$ has a probability of sampling error of at most $20$ weeks in $0.8757556$ the population mean length of time employed by all women with one job.