Chapter 10.1, Problem 14P

(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 14P

Solution: The level of significance is $\alpha =0.01$. 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 $\alpha =0.01$.

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 14P

Solution: The sampling distribution of $\overline{d}$ is Student’s t-distribution. The sample test statistic t is 1.17.

Explanation of Solution

Calculation:

	A	B	d = A - B	${(d-\overline{d})}^2$
	12.7	9.8	2.9	3.24
	13.4	14.5	-1.1	4.84
	12.8	10.7	2.1	1
	12.1	14.2	-2.1	10.24
	11.6	13	-1.4	6.25
	11.1	12.9	-1.8	8.41
	14.2	10.9	3.3	4.84
	15.1	10	5.1	16
	12.5	14.1	-1.6	7.29
	12.3	13.6	-1.3	5.76
	13.1	9.1	4	8.41
	15.8	10.2	5.6	20.25
	10.3	17.9	-7.6	75.69
	12.7	11.8	0.9	0.04
	11.1	7	4.1	9
	15.7	9.2	6.5	29.16
Sum	206.5	188.9	17.6	210.42
Average	12.90625	11.80625	1.1	13.15125

The d distribution is mound shaped and symmetrical and we have a random sample of $n=16$ paired differences, so we can assume that sample test statistics follows a Student’s t-distribution with $d.f=n-1=15$.

$\begin{array}{l}{s}_d=\sqrt{\frac{{\displaystyle \sum {(d-\overline{d})}^2}}{n-1}}\\ s_d=\sqrt{\frac{210.42}{16-1}}\\ s_d=3.7454\end{array}$

Using ${\mu }_d=0,n=16,s_d=3.7454,\overline{d}=1.1$. The test statistic t is

$\begin{array}{l}t=\frac{(\overline{d}-{\mu }_d)}{\frac{s_d}{\sqrt{n}}}\\ t=\frac{(1.1-0)}{\frac{3.7454}{\sqrt{16}}}\\ t=1.1748\\ t\approx 1.17\end{array}$

(c)

To determine

To find: The P-value of the test statistic and sketch the sampling distribution showing the area corresponding to the P-value.

Answer to Problem 14P

Solution: The P-value of the test statistic is 0.2602.

Explanation of Solution

Calculation:

We have t = 1.17

$P-\text{value}\text{=}P(t<-1.17)+P(t>1.17)$

Using Table 4 from the Appendix to find the specified area:

$0.25<P-\text{value}<0.50$

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 enter d.f = 15.

Step 4: Click on the Shaded area >X value.

Step 5: Enter X-value as 1.17 and select ‘Both Tail’.

Step 6: Click on OK.

The obtained distribution graph is:

$\begin{array}{l}P-\text{value}\text{= 2(0}\text{.1301)}\\ P-\text{value}\text{= }0.2602\end{array}$

(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 0.01.

Answer to Problem 14P

Solution: The P-value $>\alpha$, hence we fail to reject $H_0$. The data is not statistically significant for a level of significance of 0.01.

Explanation of Solution

The P-value of 0.2602 is greater than the level of significance ($\alpha$

) of 0.01. 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 0.01.

(e)

To determine

The interpretation for the conclusion.

Answer to Problem 14P

Solution: There is not enough evidence to conclude that there is difference in the population average birthrate and death rate in this region.

Explanation of Solution

The P-value of 0.2602 is greater than the level of significance ($\alpha$

) of 0.01. Therefore we don’t have enough evidence to reject the null hypothesis $H_0$. There is not enough evidence to conclude that there is difference in the population average birthrate and death rate in this region.