91Ó°ÊÓ

Consider the given question,

A simple random sample of 20quarters.

The significance level,$\mathrm{Î\pm }$ is 0.05.

The null hypothesis is given below,

$H_0:\mathrm{μ}=0.5$

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,

$H_a:\mathrm{μ}>0.2$

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.

If $P≤a$, then reject the null hypothesis.

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

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.

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.

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.

The null hypothesis is given below,

$H_0:\mathrm{μ}=0.5$

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,

$H_a:\mathrm{μ}>0.2$

The data provide sufficient evidence to conclude that on average the net percentage gain for jobs exceeds0.5.

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.

If $P≤\mathrm{Î\pm }$,then reject the null hypothesis.

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

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.

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.