Question

In a recent survey of gun control laws, a random sample of 564 women, 313 favored stricter gun control laws. In a random sample of 588 men, 307 favored stricter gun control laws. Can it be concluded at the .05 level of significance that a lower proportion of men favor stricter gun control than women

Answer 1

Answer:

The p-value of the test is 0.1314 > 0.05, which means that it cannot be concluded at the .05 level of significance that a lower proportion of men favor stricter gun control than women

Step-by-step explanation:

Before solving the question, we need to understand the central limit theorem, and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Test if a lower proportion of men favor stricter gun control than women:

At the null hypothesis, we test if the proportion is the same, that is, the difference of the proportions if 0. So

$H_0: p_M - p_W = 0$

At the alternative hypothesis, we test if it is less, that is, the subtraction of the proportions is negative. So

$H_1: p_M - p_W < 0$

The test statistic is:

$z = \frac{X - \mu}{s}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, and s is the standard error.

In a recent survey of gun control laws, a random sample of 564 women, 313 favored stricter gun control laws.

This means that:

$p_W = \frac{313}{564} = 0.555$

$s_W = \sqrt{\frac{0.555*0.445}{564}} = 0.0209$

In a random sample of 588 men, 307 favored stricter gun control laws.

This means that:

$p_M = \frac{307}{588} = 0.522$

$s_M = \sqrt{\frac{0.522*0.478}{588}} = 0.0206$

0 is tested at the null hypothesis:

This means that $\mu = 0$

From the samples:

$X = p_M - p_W = 0.522 - 0.555 = -0.033$

$s = \sqrt{s_M^2+s_W^2} = \sqrt{0.0206^2+0.0209^2} = 0.029$

Value of the test statistic:

$z = \frac{X - \mu}{s}$

$z = \frac{-0.033 - 0}{0.029}$

$z = -1.12$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion difference greater than 0.033, which is the p-value of z = -1.12.

Looking at the z-table, z = -1.12 has a p-value of 0.1314.

The p-value of the test is 0.1314 > 0.05, which means that it cannot be concluded at the .05 level of significance that a lower proportion of men favor stricter gun control than women