Chapter 10, Problem 46CE

a.

To determine

State the hypotheses.

Find the test statistic and p-value.

Interpret the results.

Answer to Problem 46CE

Hypotheses:

$H_0:{\mu }_1-{\mu }_2\le 0$

$H_1:{\mu }_1-{\mu }_2>0$

The test statistic is 0.484.

The p-value is 0.3142.

There is no evidence to infer that the first group is better than the second group.

Explanation of Solution

Calculation:

In a sample of first 25 papers, 10 received a B or better and in a sample of last 24 papers, 8 received a B or better.

Here, the claim is that first group is better than second group.

Here, the direction of the test is right-tailed.

Assume ${\pi }_1$ is the population proportion of first group and ${\pi }_2$ is the population proportion of second group.

State the hypotheses:

Null hypothesis:

$\underset{\_}{H_0:{\pi }_1-{\pi }_2\le 0}$ .

That is,the population proportion of first group is not greater than the population proportion of second group.

Alternative hypothesis:

$\underset{\_}{H_1:{\pi }_1-{\pi }_2>0}$.

That is, the population proportion of first group is greater than the population proportion of second group.

Sketch the decision rule:

Software procedure:

Step-by-step software procedure to obtain critical value using Megastat is as follows:

• • In EXCEL, Select Add-Ins>MegaStat>Probability.
• • Choose Continuous probability distributions.
• • Select Normal distribution and select calculate z given P and enter P as 0.10.
• • Enter mean as 0 and standard deviation as 1.
• • In shading, selectUpper-tail.
• • Click Ok.
• Output using Megastat software is given below:

From the output, the critical value is +1.28.

Decision rule:

If $z_{\text{calc}}>+1.28$, then reject the null hypothesis.

Sample proportion for first group:

$\begin{array}{c}{p}_1=\frac{x_1}{n_1}\\ =\frac{10}{25}\\ =0.4\end{array}$

Thus, the sample proportion for first group is 0.4.

Sample proportion for second group:

$\begin{array}{c}{p}_2=\frac{x_2}{n_2}\\ =\frac{8}{24}\\ =0.3333\end{array}$

Thus, the sample proportion for second group is 0.3333.

Pooled proportion:

$\begin{array}{c}{p}_c=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{10+8}{25+24}\\ =\frac{18}{49}\\ =0.3673\end{array}$

Thus, the pooled proportion is 0.3673.

Test statistic:

$\begin{array}{c}{z}_{\text{calc}}=\frac{p_1-p_2}{\sqrt{p_c\left(1-p_c\right)\left[\frac{1}{n_1}+\frac{1}{n_2}\right]}}\\ =\frac{0.4-0.3333}{\sqrt{0.3673\left(1-0.3673\right)\left[\frac{1}{25}+\frac{1}{24}\right]}}\\ =\frac{0.0667}{\sqrt{0.3673\left(0.6327\right)\left[0.04+0.0417\right]}}\\ =\frac{0.0667}{\sqrt{0.2324\left(0.0817\right)}}\\ =0.484\end{array}$

Thus, the test statistic is 0.484.

p-value:

Software procedure:

Step-by-step software procedure to obtain p-value using EXCEL is as follows:

• • Open an EXCEL file.
• • In cell A1, enter the formula “=NORM.S.DIST(0.484,1)”
• Output using Excel software is given below:

From the output, the p-value for right tailed test is,

$\begin{array}{c}p\text{-value}=1-0.685807\\ =0.3142\end{array}$

Thus, the p-value is 0.3142.

Decision rule:

If $p\text{-value}<\alpha$, then reject the null hypothesis.

Conclusion for p-value method:

Here, the p-value is greater than the level of significance.

That is, $p\text{-value}\left(=0.3142\right)>\alpha \left(=0.10\right)$.

Therefore, the null hypothesis is not rejected.

Thus, there is no evidence to infer that the first group is better than the second group.

b.

To determine

Check whether the samples are large enough to assume normality of $p_1-p_2$.

Answer to Problem 46CE

No, the samples are not large enough to assume normality of $p_1-p_2$.

Explanation of Solution

Calculation:

Rule for normality:

• • Rule 1: $n_1{p}_1\ge 10$
• • Rule 2: $n_1\left(1-p_1\right)\ge 10$
• • Rule 3: $n_2{p}_2\ge 10$
• • Rule 4: $n_2\left(1-p_2\right)\ge 10$

Check the rule:

Rule 1: $n_1{p}_1\ge 10$

$\begin{array}{c}{n}_1{p}_1=25\left(0.4\right)\\ =10\left(=10\right)\end{array}$

Rule 2: $n_1\left(1-p_1\right)\ge 10$

$\begin{array}{c}{n}_1\left(1-p_1\right)=25\left(1-0.4\right)\\ =15\left(>10\right)\end{array}$

Rule 3: $n_2{p}_2\ge 10$

$\begin{array}{c}{n}_2{p}_2=24\left(0.3333\right)\\ =7.9992\left(<10\right)\end{array}$

Rule 4: $n_2\left(1-p_2\right)\ge 10$

$\begin{array}{c}{n}_2\left(1-p_2\right)=24\left(1-0.3333\right)\\ =16.0008\left(>10\right)\end{array}$

Since $n_2{p}_2$ is not greater than 10, the normality of the sample proportion cannot be assumed.

c.

To determine

Make an argument that early-finishers should do better and then make the opposite argument.

Identify which is more convincing.

Explanation of Solution

Early-finishers should do better because they know the good study material and notes.Therefore, the student complete the exam faster.

The opposite argument is the students might not know about how to collect the study materials and notes. Also, the students do not know the method of study. Therefore, the student completes the exam faster.

Hence, the argument that early-finishers should do better is more convincing.