
Concept explainers
a.
State the null and alternate hypotheses.
a.

Answer to Problem 1CYU
The hypotheses are given below:
Null hypothesis:
That is, there is no significant difference between the
Alternate hypothesis:
That is, the mean reading scores of third graders after the program is greater than the mean reading scores of third graders before the program.
Explanation of Solution
The data represents the reading scores of a sample of 5 third graders before and after the reading improvement program.
Hypothesis:
Hypothesis is an assumption about the parameter of the population, and the assumption may or may not be true.
Let
Claim:
Here, the claim is, whether the mean reading scores of third graders after the program is greater than the mean reading scores of third graders before the program.
The hypotheses are given below:
Null hypothesis:
Null hypothesis is a statement which is tested for statistical significance in the test. The decision criterion indicates whether the null hypothesis will be rejected or not in the favor of alternate hypothesis.
That is, there is no significant difference between the mean reading scores of third graders before the program and mean reading scores of third graders after the program.
Alternate hypothesis:
Alternate hypothesis is contradictory statement of the null hypothesis
That is, the mean reading scores of third graders after the program is greater than the mean reading scores of third graders before the program.
b.
Compute the differences in reading scores After – Before.
b.

Answer to Problem 1CYU
The differences in reading scores After – Before is,
S.no | |
1 | 8 |
2 | 5 |
3 | –3 |
4 | 1 |
5 | 6 |
Explanation of Solution
Calculation:
The differences in reading scores After – Before is,
S.no | After | Before | |
1 | 67 | 59 | |
2 | 68 | 63 | |
3 | 78 | 81 | |
4 | 75 | 74 | |
5 | 84 | 78 |
c.
Find the value of test statistic.
c.

Answer to Problem 1CYU
The value of test statistic is 1.7318.
Explanation of Solution
Calculation:
Test statistic:
The test statistic for matched pairs is obtained as,
where
matched pairs and
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 3.40 and 4.39.
The mean and standard deviation of the differences is 3.40 and 4.39.
The test statistic is obtained as follows,
Thus, the test statistic is 1.7318.
d.
Find the P-value for the test statistic.
d.

Answer to Problem 1CYU
The P-value for the test statistic is 0.07917.
Explanation of Solution
Calculation:
Degrees of freedom:
The degrees of freedom for the test statistic is,
Thus, the degree of freedom is 4.
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 4.
- Click the Shaded Area tab.
- Choose X value and Right Tail for the region of the curve to shade.
- In X-value enter 1.7318.
- Click OK.
Output using the MINITAB software is given below:
From the MINITAB output, the P-value is 0.07917.
Thus, the P-value is 0.07917.
e.
Interpret the P-value at the level of significance
e.

Answer to Problem 1CYU
There is not enough evidence to reject the null hypothesis
Explanation of Solution
From part (d), the P-value is 0.07917.
Decision rule based on P-value:
If
If
Here, the level of significance is
Conclusion based on P-value approach:
The P-value is 0.07917 and
Here, P-value is greater than the
That is,
By the rejection rule, fail to reject the null hypothesis.
Thus, there is not enough evidence to reject the null hypothesis
f.
State the conclusion.
f.

Answer to Problem 1CYU
There is not enough evidence to conclude that the reading scores have been increased after the reading program.
Explanation of Solution
From part (e), it is known that the null hypothesis is not rejected.
That is, there is no significant difference between the mean reading scores of third graders before the program and mean reading scores of third graders after the program.
Thus, there is not enough evidence to conclude that the reading scores have been increased after the reading program.
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Chapter 9 Solutions
Essential Statistics
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