Q. 9.89 - Elementary Statistics

Question

In the article "Business Employment Dynamics: New data on Gross Job Gains and Losses", 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 for jobs as presented on the WeissStats site. Use the technology of your choice to do the following.

Part (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.

Part (b): Obtain a normal probability plot, boxplot, histogram and stem-and-leaf diagram of the data.

Part (c): Remove the outliers from the data and then repeat part (a).

Part (d): Comment on the advisability of using the z-test here.

Step-by-Step Solution

Answer

Part (a): The value of the test statistics is 1.01 and the P-value is 0.156. Hence, the data does not provide sufficient evidence to conclude that on average the net percentage gain for jobs exceeds 0.5. 

Part (b): On constructing a normal probability plot, we get,

On constructing a boxplot,

On constructing a histogram,

On constructing a stem-and leaf diagram,

Part (c): The value of the test statistics is 1.75 and the P-value is 0.04. Hence, the data provide sufficient evidence to conclude that on average the net percentage gain for jobs exceeds 0.5. 

Part (d): From part (b) there is one outlier $\left(=-1.1\right)$appears in the data.

From part (c), the z-test is calculate after removing the outlier, which provides the test statics 1.01  to 1.75.

Part (a) Step 1. Given information.

Consider the given question,

A simple random sample of 20 quarters.

The significance level,$\alpha$ is 0.05.

Part (a) Step 2. State the null and alternative hypothesis.

The null hypothesis is given below,

$H_0:\mu =0.5$

The data does not provide sufficient evidence to conclude that on average the net percentage gain for jobs exceeds 0.5.

The alternative hypothesis is given below,

$H_a:\mu >0.2$

The data provide sufficient evidence to conclude that on average the net percentage gain for jobs exceeds 0.5.

On computing the value of the test statistics,

Therefore, the value of the test statistics is 1.01 and the P-value is 0.156.

Part (a) Step 3. Interpret the result.

If $P\le a$, then reject the null hypothesis.

Here, the P-value is 0.156 which is greater than the level of significance, that is $P\left(=0.156\right)>\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is not rejected at 5% level.

Thus, it can be concluded that the results are not statistically significant at 5% level of significance.

On interpreting, we can say that the data does not provide sufficient evidence to conclude that on average the net percentage gain for jobs exceeds 0.5.

Part (b) Step 1. Construct a normal probability plot, boxplot of the data.

On constructing a normal probability plot,

From the probability plot, the observations are closer to straight line with one outlier.

On constructing a boxplot,

From the boxplot, it is clear that the distribution of gain is left skewed with one outlier.

Part (b) Step 2. Construct a histogram and stem-and-leaf diagram of the data.

On constructing a histogram,

From the histogram, it is clear that the distribution of gain is left skewed with one outlier.

On constructing a stem-and leaf diagram,

Steam-and left of GAIN $N=20$

Leaf Unit$=0.1$

From the stem-and leaf diagram, it is clear that the shape of the distribution is left skewed.

Part (c) Step 1. State the null and alternative hypothesis and remove the outliers.

The null hypothesis is given below,

$H_0:\mu =0.5$

The data does not provide sufficient evidence to conclude that on average the net percentage gain for jobs exceeds 0.5.

The alternative hypothesis is given below,

$H_a:\mu >0.2$

The data provide sufficient evidence to conclude that on average the net percentage gain for jobs exceeds 0.5.

On computing the value of the test statistics,

Test of mu$=0.2$ vs $>0.2$

The assumed standard deviation is 0.42.

Therefore, the value of the test statistics is 1.75 and the P-value is 0.04.

Part (c) Step 2. Interpret the result.

If $P\le \alpha$,then reject the null hypothesis.

Here, the P-value is 0.04 which is less than the level of significance, that is $P\left(=0.04\right)>\alpha \left(=0.05\right)$

Therefore, the null hypothesis is rejected at 5% level.

Thus, it can be concluded that the results are statistically significant at 5% level of significance.

On interpreting, we can say that the data provide sufficient evidence to conclude that on average the net percentage gain for jobs exceeds 0.5.

Part (d) Step 1. Comment on the advisability of using the z -test here.

Consider the results, it is clear that the sample size for the original data is 20. 

From part (b) there is one outlier$\left(=-1.1\right)$ appears in the data.

From part (c), the z-test is calculate after removing the outlier, which provides the test statics 1.01  to 1.75.

Hence, the z-test is calculated after removing the outlier change the conclusion of the original data.