Chapter 8.4, Problem 14P

(a)

To determine

Find the level of significance.

State the null and alternative hypothesis.

Identify the tail of test.

Answer to Problem 14P

The level of significance is 0.01.

The null hypothesis is $H_0:{\mu }_d=0$.

The alternative hypothesis is $H_1:{\mu }_d\ne 0$.

The tail of the test is two-tailed.

Explanation of Solution

Calculation:

Let ${\mu }_d$ denotes the population mean of the differences.

From the given information the value of $\alpha$ is 0.01, and the data indicate a difference between population average birth rate and death rate in this region.

Hence, the level of significance is 0.01.

The null and alternative hypothesis is,

Null hypothesis:

$H_0:{\mu }_d=0$

Alternative hypothesis:

$H_1:{\mu }_d\ne 0$

In this situation the alternative hypothesis is not equal indicates that the test is two-tailed test.

Hence, the tail of the test is two-tailed test.

(b)

To determine

Identify the sampling distribution to be used.

Mention the assumption to test.

Find the value of the sample test statistic.

Find the t value.

Answer to Problem 14P

The sampling distribution to be used is Student’s t distribution.

The Student’s t distribution is chosen because the distribution of differences is approximately normal.

The value of the sample test statistic is 1.1.

The t value is 1.175.

Explanation of Solution

Calculation:

Conditions:

• When the d distribution considered in the study has the normal distribution or simply has a mound-shaped, symmetric distribution then the sampling distribution $\overline{d}$ has Student’s t distribution for any sample size n. The t statistic is used for testing.
• When the d distribution considered in the study is not normally distributed then the sampling distribution $\overline{d}$ has Student’s t distribution if the sample size n is greater than or equal 30. That is, $n\ge 30$.

Test statistic for t:

The t statistic value for sample test statistic $\overline{d}$ is,

$t=\frac{\overline{d}-{\mu }_d}{\left(\frac{s_d}{\sqrt{n}}\right)}$

In the formula $\overline{d}$ is sample test statistic, ${\mu }_d$ is specified in $H_0$, $s_{\sigma }$ is sample standard deviation of the differences d, and n is the number of data pairs.

The distribution of differences is approximately normal is assumed.

It is clear that the distribution of differences is approximately normal. Hence, the sampling distribution to be used is Student’s t distribution.

Difference:

A	12.7	13.4	12.8	12.1	11.6	11.1	14.2	15.1
B	9.8	14.5	10.7	14.2	13.0	12.9	10.9	10.0
$d=A-B$	2.9	–1.1	2.1	–2.1	–1.4	–1.8	3.3	5.1
A	12.5	12.3	13.1	15.8	10.3	12.7	11.1	15.7
B	14.1	13.6	9.1	10.2	17.9	11.8	7.0	9.2
$d=A-B$	–1.6	–1.3	4.0	5.6	–7.6	0.9	4.1	6.5

Mean and standard deviation for difference:

Step by step procedure to obtain mean and standard deviation using MINITAB software is given as,

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the column d.
• Choose option statistics, and select Mean, standard deviation.
• Click OK.

Output using MINITAB software is given below:

From MINITAB output, the mean of the differences d is 1.10, and standard deviation of the differences d is 3.745.

The value of the sample test statistic is 1.1.

T-statistic:

Substitute $\overline{d}$ as 1.10, ${\mu }_d$ as 0, $s_{\sigma }$ as 3.745, and n as 16 in the test statistic formula

$\begin{array}{c}t=\frac{1.10-0}{\left(\frac{3.745}{\sqrt{16}}\right)}\\ =\frac{1.10}{0.93625}\\ =1.175\end{array}$

Hence, the t-value of the sample test statistic is 1.175.

(c)

To determine

Find the P-value.

Draw the sampling distribution by showing the area corresponding to the P-value.

Answer to Problem 14P

The P-value is 0.2584.

Explanation of Solution

Calculation:

Degrees of freedom:

The degrees of freedom for the t distribution is,

$d.f.=n-1$

In the formula n is the number of data pairs.

Substitute n as 16 in the degrees of freedom formula

$\begin{array}{c}d.f.=n-1\\ =16-1\\ =15\end{array}$

Step by step procedure to obtain P-value using MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• Enter the Degrees of freedom as 15.
• Click the Shaded Area tab.
• Choose X Value and Both Tails, for the region of the curve to shade.
• Enter the X value as 1.175.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the P-value is 0.1292 which is one sided value.

The two-tailed P-value is,

$\begin{array}{c}P\text{-value}=2\times 0.1292\\ =0.2584\end{array}$

Hence, the P-value is 0.2584.

(d)

To determine

Check whether the null hypothesis is rejecting or fail to reject.

Identify whether the data statistically significant at level 0.01 or not.

Answer to Problem 14P

The null hypothesis is not rejected.

The data is not statistically significant at level 0.01.

Explanation of Solution

Calculation:

From part (c), the P-value is 0.2584.

Rejection rule:

• If the P-value is less than or equal to $\alpha$, then reject the null hypothesis and the test is statistically significant. That is, $P\text{-value}\le \alpha$.

Conclusion:

The P-value is 0.2584 and the level of significance is 0.01.

The P-value is greater than the level of significance.

That is, $0.2584\left(=P\text{-value}\right)>0.01\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is not rejected.

Hence, the data is not statistically significant at level 0.01.

(e)

To determine

Interpret the conclusion in the context of the application.

Explanation of Solution

Calculation:

From part (d), the null hypothesis is not rejected. This shows that, there is no evidence that the data indicate a difference between population average birth rate and death rate in this region at level of significance 0.01.