a.
To identify: The claim and state
a.
![Check Mark](/static/check-mark.png)
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.
![Check Mark](/static/check-mark.png)
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.
![Check Mark](/static/check-mark.png)
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.
![Check Mark](/static/check-mark.png)
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.
![Check Mark](/static/check-mark.png)
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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Chapter 9 Solutions
ALEKS 360 BLUMAN ELE.STAT:A STEP.(11WKS)
- Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw HillCalculus For The Life SciencesCalculusISBN:9780321964038Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.Publisher:Pearson Addison Wesley,College Algebra (MindTap Course List)AlgebraISBN:9781305652231Author:R. David Gustafson, Jeff HughesPublisher:Cengage Learning
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