
a.
To identify: The claim and state
a.

Answer to Problem 15E
The claim is that “the meansare equal”.
The hypotheses are given below:
Null hypothesis:
Alternative hypothesis:
Explanation of Solution
Given info:
Justification:
Here, the means are equalis tested. Hence, the claim is that the means are equal. This can be written as
The hypotheses are given below:
Null hypothesis:
Alternative hypothesis:
b.
To find: The P-value.
b.

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,
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 find: The test value.
c.

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 make: The decision.
d.

Answer to Problem 15E
The decision is, the null hypothesis is rejected.
Explanation of Solution
Justification:
Decision rule:
If
If
Here, the P-value is lesser than the level of significance.
That is,
By the decision rule, the null hypothesis is rejected.
e.
To summarize: The result.
To construct: The 90% confidence interval for the difference of the mean.
e.

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
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
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