Chapter 10.5, Problem 22SE

a.

To determine

Find the sample proportions and test statistic for equal proportions.

Find the p-value.

Answer to Problem 22SE

Sample proportions: $p_1=0.4$ and $p_2=0.3$.

The test statistic is 1.4825.

The conclusion is that, there is no significant difference in the proportion of dissatisfied workers in two companies.

Explanation of Solution

Calculation:

The given information is that, $x_1=40$, $n_1=100$, $x_2=30$ and $n_2=100$

State the hypotheses:

Null hypothesis:

$H_0:{\pi }_1-{\pi }_2=0$

That is, there is no significant difference in the proportion of dissatisfied workers in two companies.

Alternative hypothesis:

$H_1:{\pi }_1-{\pi }_2\ne 0$

That is, there is a significant difference in the proportion of dissatisfied workers in two companies.

Sample proportions:

First sample:

$\begin{array}{c}{p}_1=\frac{x_1}{n_1}\\ =\frac{40}{100}\\ =0.4\end{array}$

Second sample:

$\begin{array}{c}{p}_2=\frac{x_2}{n_2}\\ =\frac{30}{100}\\ =0.3\end{array}$

Pooled proportion:

$\begin{array}{l}{p}_c=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{40+30}{100+100}\\ =\frac{70}{200}\\ =0.35\end{array}$

Thus, the pooled proportion is 0.35.

Test statistic:

$\begin{array}{c}{z}_{\text{calc}}=\frac{p_1-p_2}{\sqrt{p_c\left(1-p_c\right)\left[\frac{1}{n_1}+\frac{1}{n_2}\right]}}\\ =\frac{0.4-0.3}{\sqrt{0.35\left(1-0.35\right)\left[\frac{1}{100}+\frac{1}{100}\right]}}\\ =\frac{0.1}{\sqrt{0.2275\left(0.02\right)}}\\ =\frac{0.1}{\sqrt{0.00455}}\end{array}$

     $\begin{array}{c}=\frac{0.1}{0.067454}\\ =1.4825\end{array}$

Thus, the test statistic is 1.4825.

p-value:

$\begin{array}{c}p\text{-value}=2\left(1-\text{NORM}\text{.S}\text{.DIST}\left(z\right)\right)\\ =2\left(1-\text{NORM}\text{.S}\text{.DIST}\left(1.4825\right)\right)\end{array}$

Software procedure:

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

• Open an EXCEL file.
• In cell A1, enter the formula “=NORM.S.DIST(1.4825)”
• Output using Excel software is given below:

The p-value is,

$\begin{array}{c}2\left(1-\text{NORM}\text{.S}\text{.DIST}\left(1.4825\right)\right)=2\left(1-0.930896\right)\\ =0.1382\end{array}$

Thus, the p-value is 0.1382.

Decision rule:

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

Conclusion:

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

That is, $p\text{-value}\left(=0.1382\right)>\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is not rejected.

Thus, it can be concluded that there is no significant difference in the proportion of dissatisfied workers in two companies.

b.

To determine

Find the sample proportions and test statistic for equal proportions.

Find the p-value.

Answer to Problem 22SE

Sample proportions: $p_1=0.12$ and $p_2=0.24$.

The test statistic is –2.16.

The conclusion is that, the proportion of rooms rented at least a week in advance at the 2^nd hotel is not less than the 1^st hotel.

Explanation of Solution

Calculation:

The given information is that, $x_1=24$, $n_1=200$, $x_2=12$ and $n_2=50$

State the hypotheses:

Null hypothesis:

$H_0:{\pi }_1-{\pi }_2\ge 0$

That is, the proportion of rooms rented at least a week in advance at the 2^nd hotel is not less than the 1^st hotel.

Alternative hypothesis:

$H_1:{\pi }_1-{\pi }_2<0$

That is, the proportion of rooms rented at least a week in advance at the 2^nd hotel is less than the 1^st hotel.

Sample proportions:

First sample:

$\begin{array}{c}{p}_1=\frac{x_1}{n_1}\\ =\frac{24}{200}\\ =0.12\end{array}$

Second sample:

$\begin{array}{c}{p}_2=\frac{x_2}{n_2}\\ =\frac{12}{50}\\ =0.24\end{array}$

Pooled proportion:

$\begin{array}{l}{p}_c=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{24+12}{200+50}\\ =\frac{36}{250}\\ =0.144\end{array}$

Thus, the pooled proportion is 0.144.

Test statistic:

$\begin{array}{c}{z}_{\text{calc}}=\frac{p_1-p_2}{\sqrt{p_c\left(1-p_c\right)\left[\frac{1}{n_1}+\frac{1}{n_2}\right]}}\\ =\frac{0.12-0.24}{\sqrt{0.144\left(1-0.144\right)\left[\frac{1}{200}+\frac{1}{50}\right]}}\\ =\frac{-0.12}{\sqrt{0.1233\left(0.025\right)}}\\ =\frac{-0.12}{\sqrt{0.0030825}}\end{array}$

$\begin{array}{c}=\frac{-0.12}{0.0555}\\ =-2.16\end{array}$

Thus, the test statistic is –2.16.

p-value:

$\begin{array}{c}p\text{-value}=\text{NORM}\text{.S}\text{.DIST}\left(z\right)\\ =\text{NORM}\text{.S}\text{.DIST}\left(-2.16\right)\end{array}$

Software procedure:

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

• Open an EXCEL file.
• In cell A1, enter the formula “=NORM.S.DIST(–2.16)”
• Output using Excel software is given below:

Thus, the p-value is 0.0154.

Decision rule:

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

Conclusion:

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

That is, $p\text{-value}\left(=0.0154\right)>\alpha \left(=0.01\right)$.

Therefore, the null hypothesis is not rejected.

Thus, it can be concluded that the proportion of rooms rented at least a week in advance at the 2^nd hotel is not less than the 1^st hotel.

c.

To determine

Find the sample proportions and test statistic for equal proportions.

Find the p-value.

Answer to Problem 22SE

Sample proportions: $p_1=0.075$ and $p_2=0.05$.

The test statistic is 1.6379.

The conclusion is that, the proportion of home equity loan default rates for first bank is not greater than the second bank.

Explanation of Solution

Calculation:

The given information is that, $x_1=36$, $n_1=480$, $x_2=26$ and $n_2=520$

State the hypotheses:

Null hypothesis:

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

That is, the proportion of home equity loan default rates for first bank is not greater than the second bank.

Alternative hypothesis:

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

That is, the proportion of home equity loan default rates for first bank is greater than the second bank.

Sample proportions:

First sample:

$\begin{array}{c}{p}_1=\frac{x_1}{n_1}\\ =\frac{36}{480}\\ =0.075\end{array}$

Second sample:

$\begin{array}{c}{p}_2=\frac{x_2}{n_2}\\ =\frac{26}{520}\\ =0.05\end{array}$

Pooled proportion:

$\begin{array}{l}{p}_c=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{36+26}{480+520}\\ =\frac{62}{1,000}\\ =0.062\end{array}$

Thus, the pooled proportion is 0.062.

Test statistic:

$\begin{array}{c}{z}_{\text{calc}}=\frac{p_1-p_2}{\sqrt{p_c\left(1-p_c\right)\left[\frac{1}{n_1}+\frac{1}{n_2}\right]}}\\ =\frac{0.075-0.05}{\sqrt{0.062\left(1-0.062\right)\left[\frac{1}{480}+\frac{1}{520}\right]}}\\ =\frac{0.025}{\sqrt{\text{0}\text{.058156}\left(\text{0}\text{.004006}\right)}}\\ =1.6379\end{array}$

Thus, the test statistic is 1.6379.

p-value:

$\begin{array}{c}p\text{-value}=1-\text{NORM}\text{.S}\text{.DIST}\left(z\right)\\ =1-\text{NORM}\text{.S}\text{.DIST}\left(1.6379\right)\end{array}$

Software procedure:

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

• Open an EXCEL file.
• In cell A1, enter the formula “=NORM.S.DIST(1.6379)”
• Output using Excel software is given below:

The p-value is,

$\begin{array}{c}1-\text{NORM}\text{.S}\text{.DIST}\left(1.6379\right)=1-0.949279\\ =\text{0}\text{.0507}\end{array}$

Thus, the p-value is 0.0507.

Decision rule:

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

Conclusion:

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

That is, $p\text{-value}\left(=0.0507\right)>\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is not rejected.

Thus, it can be concluded that the proportion of home equity loan default rates for first bank is not greater than the second bank.