Question

test the given claim. identify the null hypothesis, alternative hypothesis, test statistic, p - value, and then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. among 2088 passenger cars in a particular region, 242 had only rear license plates. among 332 commercial trucks, 47 had only rear license plates. a reasonable hypothesis is that commercial trucks owners violate laws requiring front license plates at a higher rate than owners of passenger cars. use a 0.01 significance level to test that hypothesis.
a. test the claim using a hypothesis test
b. test the claim by constructing an appropriate confidence interval.
d. ( h_0: p_1 < p_2 )
( h_1: p_1 = p_2 )
identify the test statistic

• 1.34 (type an integer or a decimal. round to two decimal places as needed.)

identify the p - value
0.090 (type an integer or a decimal. round to three decimal places as needed.)
state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim.
fail to reject ( h_0 ). there is not sufficient evidence to support the claim that commercial trucks owners violate laws requiring front license plates at a higher rate than owners of passenger cars.
b. identify the confidence interval limits for the appropriate confidence interval. let population 1 correspond to the passenger cars and population 2 correspond to the commercial trucks. let a success be a vehicle that only has a rear license plate.
( square < p_1 - p_2 < square )
(type integers or decimals. round to three decimal places as needed.)

Explanation:

Step1: Calculate sample proportions

For passenger cars: \(n_1 = 2088\), \(x_1=242\), \(\hat{p}_1=\frac{x_1}{n_1}=\frac{242}{2088}\approx0.116\)
For commercial trucks: \(n_2 = 332\), \(x_2 = 47\), \(\hat{p}_2=\frac{x_2}{n_2}=\frac{47}{332}\approx0.142\)
The pooled proportion \(\hat{p}=\frac{x_1 + x_2}{n_1 + n_2}=\frac{242+47}{2088 + 332}=\frac{289}{2420}\approx0.12\)

Step2: Calculate the confidence interval

The formula for the confidence interval for \(p_1 - p_2\) is \((\hat{p}_1-\hat{p}_2)\pm z_{\alpha/2}\sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}\)
Since \(\alpha=0.01\), \(z_{\alpha/2}=z_{0.005} = 2.576\)
\(\hat{p}_1-\hat{p}_2=0.116 - 0.142=- 0.026\)
\(\sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}=\sqrt{\frac{0.116\times(1 - 0.116)}{2088}+\frac{0.142\times(1 - 0.142)}{332}}\)
\(=\sqrt{\frac{0.116\times0.884}{2088}+\frac{0.142\times0.858}{332}}\approx\sqrt{\frac{0.1025}{2088}+\frac{0.1218}{332}}\approx\sqrt{0.000049 + 0.000367}\approx\sqrt{0.000416}\approx0.0204\)
The margin of error \(E=z_{\alpha/2}\sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}=2.576\times0.0204\approx0.0526\)

Step3: Find the confidence interval limits

The lower limit is \((\hat{p}_1-\hat{p}_2)-E=-0.026-0.0526=-0.079\)
The upper limit is \((\hat{p}_1-\hat{p}_2)+E=-0.026 + 0.0526=0.027\)

Answer:

\(-0.079 < p_1 - p_2 < 0.027\)