Chapter 10.1, Problem 19P

(a)

To determine

The level of significance, null and alternative hypothesis & determine whether we should use a left-tailed, right-tailed, or two-tailed test.

Answer to Problem 19P

Solution: The level of significance is $\alpha =0.05$. The null hypothesis is $H_0:{\mu }_d=0$ and alternative hypothesis $H_A:{\mu }_d\ne 0$. This is a two-tailed test.

Explanation of Solution

The level of significance is defined as the probability of rejecting the null hypothesis when it is true, it is denoted by $\alpha =0.05$.

Null hypothesis $H_0:{\mu }_d=0$

Alternative hypothesis $H_A:{\mu }_d\ne 0$

Since $H_A:{\mu }_d\ne 0$, this is a two-tailed test.

(b)

To determine

To find: The sampling distribution that should be used along with assumptions and compute the value of the sample test statistic.

Answer to Problem 19P

Solution: The sampling distribution of $\overline{d}$ is Student’s t-distribution. The sample test statistic t is -0.82.

Explanation of Solution

Calculation:

	A	B	$d=A-B$	${(d-\overline{d})}^2$
	21	24	-3	0.5625
	25	23	2	18.0625
	20	25	-5	7.5625
	14	18	-4	3.0625
	-4	6	-10	60.0625
	19	4	15	297.5625
	15	21	-6	14.0625
	30	37	-7	22.5625
Sum	140	158	-18	423.5
Average	17.5	19.75	-2.25	52.9375

The d distribution is mound shaped and symmetrical and we have a random sample of $n=8$ paired differences, so we can assume that sample test statistics follows a Student’s t-distribution with $d.f=n-1=7$.

$\begin{array}{l}{s}_d=\sqrt{\frac{{\displaystyle \sum {(d-\overline{d})}^2}}{n-1}}\\ s_d=\sqrt{\frac{423.5}{8-1}}\\ s_d=7.7782\end{array}$

Using ${\mu }_d=0,n=8,s_d=7.778,\overline{d}=-2.25$. The test statistic t is

$\begin{array}{l}t=\frac{(\overline{d}-{\mu }_d)}{\frac{s_d}{\sqrt{n}}}\\ t=\frac{(-2.25-0)}{\frac{7.7782}{\sqrt{8}}}\\ t=-0.81818\\ t\approx -0.82\end{array}$

(c)

To determine

To find: The critical value of the test statistic and sketch the sampling distribution showing the critical region.

Answer to Problem 19P

Solution: The critical value of the test statisticis $t_C=\pm 2.365$.

Explanation of Solution

Calculation:

The level of significance is $\alpha =0.05$.

95% of the t-distribution curve lies between -$t_C$ and $t_C$.

$d.f=n-1=7$

Using table 4 from Appendix we get

$t_C=\pm 2.365$

Critical region is $-2.365>t_C>2.365$

Graph:

To draw the required graphs using the Minitab, follow the below instructions:

Step 1: Go to the Minitab software.

Step 2: Go to Graph > Probability distribution plot > View probability.

Step 3: Select ‘t’ and d.f = 7.

Step 4: Click on the Shaded area > Probability.

Step 5: Select ‘Both tail’ and enter Probability as 0.05.

Step 6: Click on OK.

The obtained distribution graph is:

Interpretation:

If the value of the sample test statistics lies in the critical region then we have to reject the null hypothesis.

(d)

To determine

Whether we reject or fail to reject the null hypothesis and whether the data is statistically significant for a level of significance of 0.05.

Answer to Problem 19P

Solution: The t-value (- 0.82) does not lie in the critical region. The data is not statistically significant for a level of significance of 0.05

Explanation of Solution

The critical regionis $-2.365>t_C>2.365$.

The t-value (- 0.82)does not lie in the critical region. Therefore we don’t have enough evidence to reject the null hypothesis $H_0$ hence the data is not statistically significant for a level of significance of 0.05.

(e)

To determine

The interpretation for the conclusion.

Answer to Problem 19P

Solution: There is not enough evidence to conclude that the populations mean percentage increase in corporate revenue is different from the population mean percentage increase in CEO salary.

Explanation of Solution

The t-value (- 0.82) does not lie in the critical region. Therefore we don’t have enough evidence to reject the null hypothesis $H_0$. There is not enough evidence to conclude that the populations mean percentage increase in corporate revenue is different from the population mean percentage increase in CEO salary.

(f)

To determine

The comparison for our conclusion with the conclusion obtained from the P-value method.

Answer to Problem 19P

Solution: Both the conclusions are same.

Explanation of Solution

The conclusion from the P-value method is:

The P-value of 0.4392 is greater than the level of significance ($\alpha$) of 0.05. There is not enough evidence to conclude that the populations mean percentage increase in corporate revenue is different from the population mean percentage increase in CEO salary.

Since we are using the same significance level of 0.05, hence both the conclusions will be same.