Question

Among 2165 passenger cars in a particular​ region, 235 had only rear license plates. Among 330 commercial​ trucks, 50 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 0.10 significance level. Identify the P-value. 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 Group of answer choices

Answer 1

Using the z-distribution, it is found that the p-value is of 0.0192.

At the null hypothesis, it is tested if commercial trucks owners do not violate laws requiring front license plates at a higher rate than owners of passenger cars, that is, the subtraction is of at most 0, hence:

$H_0: p_2 - p_1 \leq 0$

At the alternative hypothesis, it is tested if commercial truck owners violate the laws more, that is, the subtraction is positive, hence:

$H_1: p_2 - p_1 > 0$

For each sample, the size, the proportion and the standard error are given by:

$n_1 = 2165, p_1 = \frac{235}{2165} = 0.1085, s_1 = \sqrt{\frac{0.1085(0.8915)}{2165}} = 0.0067$

$n_2 = 330, p_1 = \frac{50}{330} = 0.1515, s_1 = \sqrt{\frac{0.1515(0.8485)}{330}} = 0.0197$

For the distribution of differences, the mean and the standard error are given by:

$\overline{p} = p_2 - p_1 = 0.1515 - 0.1085 = 0.043$

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{0.0067^2 + 0.0197^2} = 0.0208$

The test statistic is given by:

$z = \frac{\overline{p} - p}{s}$

In which $p = 0$ is the value tested at the null hypothesis.

Hence:

$z = \frac{\overline{p} - p}{s}$

$z = \frac{0.043 - 0}{0.0208}$

$z = 2.07$

The p-value is found using a z-distribution calculator, for a right-tailed test, as we are testing if the proportion is more than a value, with z = 2.07.

• Using the calculator, the p-value is of 0.0192.

A similar problem is given at brainly.com/question/15545277