Chapter 11.1, Problem 11P

(a)

To determine

Find the level of significance.

State the null and alternative hypothesis.

Answer to Problem 11P

The level of significance is 0.01.

Explanation of Solution

Calculation:

From the given information the value of $\alpha$ is 0.01, and the information indicate that the dropout rates for males and females are different.

Hence, the level of significance is 0.01.

The null and alternative hypothesis is,

Null hypothesis:

$H_0:$ The distribution of the dropout rates for males and females are same.

Alternative hypothesis:

$H_1:$ The distribution of the dropout rates for males and females are different.

(b)

To determine

Identify the sampling distribution to be used.

Find the value of the sample test statistic.

Answer to Problem 11P

The sampling distribution to be used is normal distribution.

The value of the sample test statistic is 0.

Explanation of Solution

Calculation:

Conditions:

The conditions for using the normal distribution to test a proportion p with proportion of success $\left(p\right)$ and proportion of failure $\left(q=1-p\right)$ for a sufficiently large n (number of trails) are,

• $np>5$
• $nq>5$

The sample statistic has the normal distribution with mean p, and standard deviation $\sqrt{\frac{pq}{n}}$.

Test statistic:

The z value for the sample test statistic x is,

$z=\frac{x-0.5}{\sqrt{\frac{0.25}{n}}}$

In the formula n is the total number of plus and minus signs, x is the total number of plus signs divided by n, p is proportion specified in $H_0$, and $q=1-p$.

The sign is obtained by taking the difference of males and females. If the difference is positive then assign ‘+’ to the corresponding data pair. If the difference is negative then assign ‘–’ to the corresponding data pair. If the difference is zero then remove the data pair from the calculation.

The standard normal distribution is used as the sampling distribution for the sign test.

The signs are,

Region	Male	Female	Difference	Sign of difference
1	7.3	7.5	–0.2	–
2	7.5	6.4	1.1	+
3	7.7	6.0	1.7	+
4	21.8	20.0	1.8	+
5	4.2	2.6	1.6	+
6	12.2	5.2	7.0	+
7	3.5	3.1	0.4	+
8	4.2	4.9	–0.7	–
9	8.0	12.1	–4.1	–
10	9.7	10.8	–1.1	–
11	14.1	15.6	–1.5	–
12	3.6	6.3	–2.7	–
13	3.6	4.0	–0.4	–
14	4.0	3.9	0.1	+
15	5.2	9.8	–4.6	–
16	6.9	9.8	–2.9	–
17	15.6	12.0	3.6	+
18	6.3	3.3	3.0	+
19	8.0	7.1	0.9	+
20	6.5	8.2	–1.7	–

There are ten positive signs and totally 20 positive and negative signs are there.

The value of x is,

$\begin{array}{c}x=\frac{\text{number of plus signs}}{\text{total number of signs}}\\ =\frac{10}{20}\\ =0.5\end{array}$

The value of x is 0.5.

Test statistic:

Substitute x as 0.5, and n as 20 in the test statistic formula

$\begin{array}{c}z=\frac{0.5-0.5}{\sqrt{\frac{0.25}{20}}}\\ =\frac{0}{0.1118}\\ =0\end{array}$

Hence, the z value is 0.

(c)

To determine

Find the P-value of the sample test statistic.

Answer to Problem 11P

The P-value is 1.00.

Explanation of Solution

Calculation:

Step by step procedure to obtain P-value using MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X Value and Both Tail, for the region of the curve to shade.
• Enter the X value as 0.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the P-value is 0.5 which is one sided value.

The two-tailed P-value is,

$\begin{array}{c}P\text{-value}=2\times 0.5\\ =1.00\end{array}$

Hence, the P-value is 1.00.

(d)

To determine

Mention the conclusion of the test.

Answer to Problem 11P

The null hypothesis is failed to be rejected.

Explanation of Solution

Calculation:

From part (c), the P-value is 1.00.

Rejection rule:

• If the P-value is less than or equal to $\alpha$, then reject the null hypothesis and the test is statistically significant. That is, $P\text{-value}\le \alpha$.

Conclusion:

The P-value is 1.00 and the level of significance is 0.01.

The P-value is greater than the level of significance.

That is, $1.00\left(=P\text{-value}\right)>0.01\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is failed to be rejected.

Hence, the data is not statistically significant at level 0.01.

(e)

To determine

Interpret the conclusion in the context of the application.

Explanation of Solution

Calculation:

From part (d), the null hypothesis is failed to be rejected. This shows that, there is no sufficient evidence that the distribution of the dropout rates for males and females are different at level of significance 0.01.