9.89 Job Gains and Losses. In the article "Business Employment Dynamics: New Data on Gross Job Gains and Losses" (Monthly Labor Review, Vol. 127. Issue 4. pp. 29-42). J. Spletzer et al. examined gross job gains and losses as a percentage of the average of previous and current employment figures. A simple random sample of 20 quarters provided the net percentage gains (losses are negative gains) for jobs as presented on the WeissStats site. Use the technology of your choice to do the following.
a. Decide whether, on average, the net percentage gain for jobs exceeds 0.2. Assume a population standard deviation of 0.42. Apply the one-mean z-test with a $5\%$ significance level.
b. Obtain a normal probability plot, boxplot, histogram, and stem-and-leaf diagram of the data.
c. Remove the outliers (if any) from the data and then repeat part (a).
d. Comment on the advisability of using the z-test here.

Let, the population standard deviation is $0.42$. To decide whether, on average, the net percentage gain for jobs exceeds $0.2$.

Let the null hypothesis as follows:
$H_0:\mu =0.5$
The information does not show that the net percentage growth in jobs is greater than $0.5$ on average.
Then the alternative hypothesis
$H_0:\mu >0.2$
The results show that, on average, the net percentage gain in jobs is greater than $0.5$.

And the significance level is $\alpha =0.05$
MINITAB can be used to calculate the test statistic and P-value.

The test statistic value is $1.01$, and the P-value is $0.156$, according to the MINITAB output.

If $P\le \alpha$, the null hypothesis must be rejected.
The P-value is $0.156$, which is higher than the significance level.
$P(=0.156)>\alpha (=0.05)$
At the $5$percent level, the null hypothesis is not rejected.

As a result, the data does not support the sufficient evidence to conclude that the net percentage gain for jobs is greater than $0.5$on average.

To obtain a normal probability plot, boxplot, histogram, and stem-and-leaf diagram of the data.

The net percentage gain for job was calculated using a random sample of $20$ quarters, as shown on the Weiss Stats site.
MINITAB is used to create the normal probability plot.
The output of MINITAB will be:

With one outlier, the observations appear to be closer to a straight line on the probability map.
MINITAB is used to create the boxplot.
The output of MINITAB will be:

The boxplot clearly shows that the gain distribution is skewed due to one outlier.
MINITAB is used to create the histogram.
The output of MINITAB will be:

Using MINITAB, create the stem-and-leaf diagram.
The output of MINITAB will be:
Stem-and-leaf Display: GAIN

The shape of the distribution is left skewed, as shown in the stem and leaf diagram.
As a result, the distribution's shape is left skewed.

To remove the outliers from the data and then repeat part (a).

Let, the null hypothesis as follows:
$H_0:\mu =0.5$
The evidence does not show that the net percentage gain in jobs is greater than $0.5$ on average.
And the alternative hypothesis as follows:
$H_0:\mu >0.2$
The analysis shows that, on average, the net percentage gain in jobs is greater than $0.5$.
The significance level is $\alpha =0.05$.
MINITAB can be used to calculate the test statistic and P-value.

The output MINITAB will be:
One sample Z: Gain

The test statistic value is $1.75$, and the P-value is $0.040$, according to the MINITAB report.
If $P\le \alpha$, the null hypothesis must be rejected.
The P-value in this case is $0.040$, which is less than the significance level.
$P(=0.040)\le \alpha (=0.05)$
At the $5\%$ level, the null hypothesis is rejected.
As a result, the data give sufficient evidence to indicate that the net percentage gain for jobs is greater than $0.5$ on average.

To comment on the advisability of using the $z$-test here.

The net percentage gain for job was calculated using a random sample of $20$quarters, as shown on the Weiss Stats site.
The sample size for the original data is clearly $20$based on the preceding results.
One outlier $(=-1.1)$ emerges in the data from part (b).
After removing the outlier in part (c), the z-test is calculated, yielding the test statistic.
$1.01$to $1.75$.
As a result, after eliminating the outlier charge from the original data, the $z$-test is calculated.