a.
Compute the differences in samples of matched pairs Sample-1– Sample-2.
a.
Answer to Problem 10E
The differences in samples of matched pairs Sample-1– Sample-2 is,
S.no | Difference |
1 | –6 |
2 | –1 |
3 | –9 |
4 | –1 |
5 | –5 |
6 | –1 |
7 | –4 |
8 | –8 |
9 | –2 |
10 | –9 |
Explanation of Solution
Calculation:
The data represents the sample of a five matched pairs.
Let
The differences in samples of matched pairs Sample-1– Sample-2 is,
S.no | Sample-1 | Sample-2 | |
1 | 28 | 34 | |
2 | 29 | 30 | |
3 | 22 | 31 | |
4 | 25 | 26 | |
5 | 26 | 31 | |
6 | 29 | 30 | |
7 | 27 | 31 | |
8 | 24 | 32 | |
9 | 27 | 29 | |
10 | 28 | 37 |
b.
Find the value of t-test statistic.
b.
Answer to Problem 10E
The value of test statistic is –4.3948.
Explanation of Solution
Calculation:
The test hypotheses are given below:
Null hypothesis:
That is, there is no significant difference between the population mean1 and population mean2.
Alternate hypothesis:
That is, there is a significant difference between the population mean1 and population mean2.
Mean and standard deviation of differences:
Software procedure:
Step-by-step procedure to obtain the mean and standard deviation using the MINITAB software:
- Choose Stat > Basic statistic > Display
descriptive statistics . - In Variables, enter the column of Differences.
- In Statistics, select mean, standard deviation and N total.
- Click OK.
Output using the MINITAB software is given below:
From the MINITAB output, the mean and standard deviation are –4.6 and 3.31.
The mean and standard deviation of the differences is –4.6 and 3.31.
The test statistic is obtained as follows,
Thus, the test statistic is –4.3948.
c.
Check whether the null hypothesis is rejected at
c.
Answer to Problem 10E
There is enough evidence to reject the null hypothesis
Explanation of Solution
Degrees of freedom:
The degrees of freedom for the test statistic is,
Thus, the degree of freedom is 9.
P-value:
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 9.
- Click the Shaded Area tab.
- Choose X value and Both Tail for the region of the curve to shade.
- In X-value enter –4.3948.
- Click OK.
Output using the MINITAB software is given below:
From the MINITAB output, the P-value is
Thus, the P-value is 0.0017.
Decision rule based on P-value:
If
If
Here, the level of significance is
Conclusion based on P-value approach:
The P-value is 0.0017 and
Here, P-value is less than the
That is,
By the rejection rule, reject the null hypothesis.
Thus, there is enough evidence to reject the null hypothesis
d.
Check whether the null hypothesis is rejected at
d.
Answer to Problem 10E
There is enough evidence to reject the null hypothesis
Explanation of Solution
From part (c), the P-value is 0.0017.
Decision rule based on P-value:
If
If
Here, the level of significance is
Conclusion based on P-value approach:
The P-value is 0.0017 and
Here, P-value is less than the
That is,
By the rejection rule, reject the null hypothesis.
Thus, there is enough evidence to reject the null hypothesis
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Chapter 9 Solutions
Essential Statistics
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