Chapter 9.2, Problem 15E

a.

To determine

To identify: The claim and state $H_0$ and $H_1$.

Answer to Problem 15E

The claim is that “the meansare equal”.

The hypotheses are given below:

Null hypothesis: $H_0:{\mu }_1={\mu }_2$ (claim)

Alternative hypothesis: $H_1:{\mu }_1\ne {\mu }_2$

Explanation of Solution

Given info:

${\overline{x}}_1=2.3$, ${\overline{x}}_2=1.9$, $s_1=0.6$, $s_2=0.3$, $n_1=16$, and $n_2=16$.

Justification:

Here, the means are equalis tested. Hence, the claim is that the means are equal. This can be written as ${\mu }_1={\mu }_2$. The complement of the claim is ${\mu }_1\ne {\mu }_2$. In the given experiment, the null hypothesis indicates the claim.

The hypotheses are given below:

Null hypothesis:

$H_0:{\mu }_1={\mu }_2$

Alternative hypothesis:

$H_1:{\mu }_1\ne {\mu }_2$

b.

To determine

To find: The P-value.

Answer to Problem 15E

The P-value is 0.03042.

Explanation of Solution

Calculation:

Here, variances are not equal. Hence, the degrees of freedom is,

$\begin{array}{c}\text{df}=\text{Smallar of }{n}_1-1\text{ and }{n}_2-1\\ =16-1\\ =15\end{array}$

Software procedure:

Step by step procedure to obtain the P-value using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability> OK.
• From Distribution, choose ‘t’ distribution.
• In Degrees of freedom, enter 15.
• Click the Shaded Area tab.
• Choose X Value and Two Tail for the region of the curve to shade.
• Enter the X value as 2.39.
• Click OK.

Output using the MINITAB software is given below:

From the output, the P-value is 0.03042.

c.

To determine

To find: The test value.

Answer to Problem 15E

The test value is 1.92.

Explanation of Solution

Calculation:

Test statistic:

Software Procedure:

Step by step procedure to obtain test statistic using the MINITAB software:

• Choose Stat > Basic Statistics > 2-Sample t.
• Choose Summarized data.
• In first, enter Sample size as16, Mean as 2.3, Standard deviation as 0.6.
• In second, enter Sample size as16, Mean as 1.9, Standard deviation as 0.3.
• Choose Options.
• In Confidence level, enter 99.
• In Alternative, select not equal.
• Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

From the MINITAB output, the test value is 2.39.

d.

To determine

To make: The decision.

Answer to Problem 15E

The decision is, the null hypothesis is rejected.

Explanation of Solution

Justification:

Decision rule:

If $P\text{-value}\le \alpha$, reject the null hypothesis

If $P\text{-value}>\alpha$, do not reject the null hypothesis.

Here, the P-value is lesser than the level of significance.

That is, $P\text{-value}\left(=0.03042\right)<\alpha \left(=0.01\right)$.

By the decision rule, the null hypothesis is rejected.

e.

To determine

To summarize: The result.

To construct: The 90% confidence interval for the difference of the mean.

Answer to Problem 15E

The conclusion is that, there is enough evidence to support the claim that the means are equal.

The 90% confidence interval for the difference of the mean is $\left(0.112,0.688\right)$.

Explanation of Solution

Justification:

From part (d), the null hypothesis is rejected. Thus, there is enough evidence to support the claim that the means are equal.

Confidence interval:

Software Procedure:

Step by step procedure to obtain the 90% confidence interval for the difference of the mean using the MINITAB software:

• Choose Stat > Basic Statistics > 2-Sample t.
• Choose Summarized data.
• In first, enter Sample size as16, Mean as 2.3, Standard deviation as 0.6.
• In second, enter Sample size as16, Mean as 1.9, Standard deviation as 0.3.
• Choose Options.
• In Confidence level, enter 90.
• In Alternative, select not equal.
• Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

From the MINITAB output, the 90% confidence interval for the difference of the meanis $\left(0.112,0.688\right)$.