91影视

In testing of owners who violate license plate laws, among 2049 Connecticut passenger鈥檚 cars only 239 cars had license plates, and among 334 Connecticut trucks, only 45 had license plates.

Level of significance to test hypothesis is 0.05.

a. Let $p_1$be the population proportion of car owners who violate rear license plate laws and $p_2$be the population proportion of truck owners who violate rear license plate laws.

Mathematically, the test hypothesis is

$$\begin{gathered} H_0:p_1=p_2 \\ {H}_1:p_1>p_2 \end{gathered}$$

Thus, the test is right tailed.

Test statistics is given by,

$z=\frac{\left({\hat{p}}_1-{\hat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{\frac{\stackrel{炉}{p}\stackrel{炉}{q}}{n_1}+\frac{\stackrel{炉}{p}\stackrel{炉}{q}}{n_2}}}$

Where, ${\hat{p}}_{1}and{\hat{p}}_{\text{2}}$are estimated proportions, $\stackrel{炉}{p}$is pooled proportions and $\stackrel{炉}{q}=1-\stackrel{炉}{p}$.

Also, $n_{1}and{n}_2$are sample size of for each of the sample.

Let ${\hat{p}}_1$be the estimated proportion of car owners who violate license plate laws and ${\hat{p}}_2$be the estimated proportion of truck owners who violates rear license plate laws.

Thus,

$$\begin{gathered} {\hat{p}}_1=\frac{239}{2049} \\ {\hat{p}}_1=0.1166 \\ {\hat{q}}_1=1-{\hat{p}}_1 \\ {\hat{q}}_1=0.8834 \end{gathered}$$

Also,

$$\begin{gathered} {\hat{p}}_2=\frac{45}{334} \\ {\hat{p}}_2=0.1347 \\ {\hat{q}}_2=1-{\hat{p}}_2 \\ {\hat{q}}_{}=0.8653 \end{gathered}$$

Here, $x_1$and $x_2$are number of owners who violate rear license plate laws of cars and trucks respectively.

The pooled proportions is computed as,

$$\begin{gathered} \stackrel{炉}{p}=\frac{x_1+x_2}{n_1+n_2} \\ =\frac{239+45}{2049+334} \\ =0.1192 \\ \stackrel{炉}{q}=0.8808 \end{gathered}$$

Thus,$\stackrel{炉}{p}\stackrel{炉}{q}=0.1050$

Test statistic is given by,

$$\begin{gathered} z=\frac{\left({\hat{p}}_1-{\hat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{\frac{\stackrel{炉}{p}\stackrel{炉}{q}}{n_1}+\frac{\stackrel{炉}{p}\stackrel{炉}{q}}{n_2}}} \\ =\frac{\left(0.1166-0.1347\right)}{\sqrt{\frac{0.1050}{2049}+\frac{0.1050}{334}}} \\ =-0.9461 \end{gathered}$$

If test statistic is greater than critical value, reject null hypothesis under level of significance.

Refer to z-table for the critical value. The right-tailed test implies the cumulative area to the left of the critical value would be 0.95(1-0.05).

Thus, the critical value is 1.645.

Here, the value of the test statistic is less than the critical value.

Therefore, we failed to reject the null hypothesisunder a 0.05 significance level.

There is sufficient evidence to reject the claim that passenger car owners violate license plate laws at a higher rate than owners of commercial trucks.

b. The formula for confidence interval to test the claim is about difference between two population proportion $p_1\text{and}{p}_2$is given by,

$\left({\hat{p}}_1-{\hat{p}}_2\right)-E<p_1-p_2<\left({\hat{p}}_1-{\hat{p}}_2\right)+E$

Where, E is margin of error.

The 0.05 level of significance for one-tailed test implies 90% level of confidence.

The critical value for 90% confidence level is obtained as 1.645 $\left(z_{\frac{0.10}{2}}\right)$using the standard normal table.

E is margin of error which is given by,

$$\begin{gathered} E=z_{\frac{伪}{2}}\sqrt{\frac{{\hat{p}}_1{\hat{q}}_1}{n_1}+\frac{{\hat{p}}_2{\hat{q}}_2}{n_2}} \\ =1.645脳\sqrt{\frac{0.1028}{2049}+\frac{0.1165}{334}} \\ =0.0329 \end{gathered}$$

Substitute the values for 90% confidence interval as,

$$\begin{gathered} \left({\hat{p}}_1-{\hat{p}}_2\right)-E<p_1-p_2<\left({\hat{p}}_1-{\hat{p}}_2\right)+E=(0.1164-0.1347)-0.033<p_1-p_2<(0.1164-0.1347)+0.033 \\ =-0.0510<p_1-p_2<0.0148 \end{gathered}$$

The 90% confidence interval under is.

This interval contains zero that means there is no significant difference between two proportions ${\text{p}}_{\text{1}}{\text{andp}}_{\text{2}}$.

Therefore, the confidence interval suggests that there is no sufficient evidence to support the claim thatpassenger car owners violate license plate laws at a higher rate than owners of commercial trucks.