Chapter 10, Problem 91CE

Count the number of two-door vehicles among 50 vehicles from a college or university student parking lot. Use any sampling method you like (e.g., the first 50 you see). Do the same for a grocery store that is not very close to the college or university. At α = .10, is there a significant difference in the proportion of two-door vehicles in these two locations? (a) State the hypotheses. (b) State the decision rule and sketch it. (c) Find the sample proportions and z test statistic. (d) Make a decision. (e) Find the p-value and interpret it. (f) Is the normality assumption fulfilled? Explain.

To determine

Count the number of two-door vehicles among 50 vehicles from a college or university student parking lot by using any sampling method.

Count the number of two-door vehicles among 50 vehicles from a grocery store that is not very close to the college or university by using any sampling method.

State the hypotheses.

Explanation of Solution

Answer will vary. One of the possible answers is given below:

Simple random sampling:

If a sample of n subjects is selected, then every unit in the population has the equal chance of being selected into the considered sample then it is termed as simple random sampling. In simple random sample all the units have equal chance for being in the sample and are randomly selected. That is, select the sample of items with equal chance of priority.

Sample 1: Steps to find the number of two-door vehicles by using the simple random sampling:

• • Select the 50 vehicles from a college or university student parking lot using simple random sampling.
• • On examination, it is observed that 35 vehicles have two-doors.

Therefore, the total number of vehicles is $n_1=50$ and the number of two-door vehicles is $x_1=35$.

Sample 2: Steps to find the number of two-door vehicles by using the simple random sampling:

• • Select the 50 vehicles from a grocery store that is not very close to the college or university.
• • On examination, it is observed that 20 vehicles have two-doors.

Therefore, the total number of vehicles is $n_2=50$ and the number of two-door vehicles is $x_2=20$.

Here, the claim is that there is a significant difference in the proportion of two-door vehicles in these two locations.

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

Assume ${\mu }_G$ is the population mean number of two-door vehicles in grocery store and ${\mu }_U$ is the population mean number of two-door vehicles in for university.

State the hypotheses:

Null hypothesis:

$H_0:{\mu }_U-{\mu }_G=0$

That is, there is no significant difference in the proportion of two-door vehicles in these two locations.

Alternative hypothesis:

$H_1:{\mu }_U-{\mu }_G\ne 0$

That is, there is a significant difference in the proportion of two-door vehicles in these two locations.

To determine

State the decision rule and sketch it.

Answer to Problem 91CE

The decision rule is, “reject the null hypothesis if $z_{\text{calc}}<-1.645$ or $z_{\text{calc}}>+1.645$”.

Explanation of Solution

Calculation:

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, select Two-tail.
• • Click Ok.
• Output using Megastat software is given below:

From the output, the critical values are $\pm 1.645$.

Decision rule:

If $z_{\text{calc}}>+1.645$, then reject the null hypothesis.

If $z_{\text{calc}}<-1.645$, then reject the null hypothesis.

To determine

Find the sample proportions and test statistic.

Answer to Problem 91CE

The sample proportion for first sample is 0.7 and the sample proportion for second sample is 0.4.

The test statistic is 3.015.

Explanation of Solution

Calculation:

Sample proportion for sample 1:

$\begin{array}{c}{p}_1=\frac{x_1}{n_1}\\ =\frac{35}{50}\\ =0.7\end{array}$

Thus, the sample proportion for sample 1 is 0.7.

Sample proportion for sample 2:

$\begin{array}{c}{p}_2=\frac{x_2}{n_2}\\ =\frac{20}{\text{50}}\\ =0.4\end{array}$

Thus, the sample proportion for sample 2 is 0.4.

Pooled proportion:

$\begin{array}{c}{p}_c=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{35+20}{50+50}\\ =\frac{55}{100}\\ =0.55\end{array}$

Thus, the pooled proportion is 0.55.

Test statistic:

$\begin{array}{c}{z}_{\text{calc}}=\frac{p_1-p_2}{\sqrt{p_c\left(1-p_c\right)\left[\frac{1}{n_1}+\frac{1}{n_2}\right]}}\\ =\frac{0.7-0.4}{\sqrt{0.55\left(1-0.55\right)\left[\frac{1}{50}+\frac{1}{50}\right]}}\\ =\frac{0.3}{\sqrt{0.55\left(0.45\right)\left[0.02+0.02\right]}}\\ =\frac{0.3}{\sqrt{0.2475\left(0.04\right)}}\\ =\frac{0.3}{0.0995}\\ =3.015\end{array}$

Thus, the test statistic is 3.015.

To determine

Make a decision.

Answer to Problem 91CE

The conclusion is that there is a significant difference in the proportion of two-door vehicles in these two locations.

Explanation of Solution

Here, the test statistic is greater than the critical value.

That is, $z_{\text{calc}}\left(=3.015\right)>+1.645$.

Therefore, the null hypothesis is rejected.

Thus, there is a significant difference in the proportion of two-door vehicles in these two locations.

To determine

Find the p-value and interpret it.

Answer to Problem 91CE

The p-value is 0.0026.

The conclusion is that there is a significant difference in the proportion of two-door vehicles in these two locations.

Explanation of Solution

Calculation:

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(3.015,1)”
• Output using Excel software is given below:

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

$\begin{array}{c}p\text{-value}=2\left(1-0.998715\right)\\ =0.0026\end{array}$

Decision rule:

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

Conclusion for p-value method:

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

That is, $p\text{-value}\left(=0.0026\right)<\alpha \left(=0.10\right)$.

Therefore, the null hypothesis is rejected.

Thus, there is a significant difference in the proportion of two-door vehicles in these two locations.

To determine

Check whether the normality of $p_1-p_2$ can be assumed or not.

Answer to Problem 91CE

The normality assumption is fulfilled.

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: $n_1{p}_1\ge 10$

$\begin{array}{c}{n}_1{p}_1=50\left(0.7\right)\\ =35\left(>10\right)\end{array}$

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

$\begin{array}{c}{n}_1\left(1-p_1\right)=50\left(1-0.7\right)\\ =15\left(>10\right)\end{array}$

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

$\begin{array}{c}{n}_2{p}_2=50\left(0.4\right)\\ =20\left(>10\right)\end{array}$

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

$\begin{array}{c}{n}_2\left(1-p_2\right)=50\left(1-0.4\right)\\ =30\left(>10\right)\end{array}$

Since $n_1{p}_1$, $n_1\left(1-p_1\right)$, $n_2{p}_2$ and $n_2\left(1-p_2\right)$ are greater than 10, the normality of the sample proportion can be assumed.