Chapter 9.3, Problem 1CYU

a.

To determine

State the null and alternate hypotheses.

Answer to Problem 1CYU

The hypotheses are given below:

Null hypothesis:

$H_0:{\mu }_d=0$

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:

$H_1:{\mu }_d>0$

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 ${\mu }_d$ be the population mean difference in reading scores after and before the program (After – Before).

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.

$H_0:{\mu }_d=0$

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

$H_1:{\mu }_d>0$

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.

To determine

Compute the differences in reading scores After – Before.

Answer to Problem 1CYU

The differences in reading scores After – Before is,

S.no	$\text{Difference}$
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	$\text{Difference}\text{=}\text{After}-\text{Before}$
1	67	59	$67-59=8$
2	68	63	$68-63=5$
3	78	81	$78-81=-3$
4	75	74	$75-74=1$
5	84	78	$84-78=6$

c.

To determine

Find the value of test statistic.

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,

$t=\frac{\overline{d}-{\mu }_0}{\left(\frac{s_d}{\sqrt{n}}\right)}$,

where $\overline{d}$ is the sample mean of the differences between the values in the matched pairs, $s_d$ is the sample standard deviation of the differences between the values in the

matched pairs and ${\mu }_d$ is the population mean difference for matched pair.

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,

$\begin{array}{c}t=\frac{\overline{d}-0}{\left(\frac{s_d}{\sqrt{n}}\right)}\\ =\frac{3.40-0}{\left(\frac{4.39}{\sqrt{5}}\right)}\\ =\frac{3.40}{1.9633}\\ =1.7318\end{array}$

Thus, the test statistic is 1.7318.

d.

To determine

Find the P-value for the test statistic.

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,

$\begin{array}{c}d.f=n-1\\ =5-1\\ =4\end{array}$

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.

To determine

Interpret the P-value at the level of significance $\alpha =0.05$.

Answer to Problem 1CYU

There is not enough evidence to reject the null hypothesis $H_0$.

Explanation of Solution

From part (d), the P-value is 0.07917.

Decision rule based on P-value:

If $P-\text{value}\le \alpha$, then reject the null hypothesis $H_0$.

If $P-\text{value}>\alpha$, then fail to reject the null hypothesis $H_0$.

Here, the level of significance is $\alpha =0.05$.

Conclusion based on P-value approach:

The P-value is 0.07917 and $\alpha$ value is 0.05.

Here, P-value is greater than the $\alpha$ value.

That is, $0.07917\left(=P-\text{value}\right)>0.05\left(=\alpha \right)$.

By the rejection rule, fail to reject the null hypothesis.

Thus, there is not enough evidence to reject the null hypothesis $H_0$.

f.

To determine

State the conclusion.

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.