Question

A survey of 1100 adults from a certain region​ asked, "If purchasing a used car made certain upgrades or features more​ affordable, what would be your preferred luxury​ upgrade?" The results indicated that 49​% of the females and 41​% of the males answered window tinting. The sample sizes of males and females were not provided. Suppose that of 600 ​females, 294 reported window tinting as their preferred luxury upgrade of​choice, while of 500 ​males, 205 reported window tinting as their preferred luxury upgrade of choice. Complete parts​ (a) through​ (d) below.
a. Is there evidence of a difference between males and females in the proportion who said they prefer window tinting as a luxury upgrade at the 0.05
level of​ significance?
b. State the null and alternative​hypotheses, where π1 is the population proportion of females who said they prefer window tinting as a luxury upgrade and π2 is the population proportion of males who said they prefer window tinting as a luxury upgrade.

Answer 1

Answer:

a) The p-value of the test is 0.0076 < 0.05, which means that there is evidence of a difference between males and females in the proportion who said they prefer window tinting as a luxury upgrade at the 0.05.

b) The null hypothesis is $H_0: \pi_1 - \pi_2 = 0$ and the alternate hypothesis is $H_1: \pi_1 - \pi_2 \neq 0$.

Step-by-step explanation:

Before testing the hypothesis, 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.

Females:

49% from a sample of 600. So

$\pi_1 = 0.49, s_{\pi_1} = \sqrt{\frac{0.49*0.51}{600}} = 0.0204$

Males:

41% from a sample of 500. So

$\pi_2 = 0.41, s_{\pi_2} = \sqrt{\frac{0.41*0.59}{500}} = 0.022$

Test if there is a difference between males and females in the proportion who said they prefer window tinting as a luxury upgrade.

From here, question b can already be answered.

At the null hypothesis we test if there is no difference, that is, the subtraction of the proportions is 0. So

$H_0: \pi_1 - \pi_2 = 0$

At the alternate hypothesis, we test if there is a difference, that is, the subtraction of the proportions is different of 0. So

$H_1: \pi_1 - \pi_2 \neq 0$

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.

0 is tested at the null hypothesis:

This means that $\mu = 0$

From the two samples:

$X = \pi_1 - \pi_2 = 0.49 - 0.41 = 0.08$

$s = \sqrt{s_{\pi_1}^2 + s_{\pi_2}^2} = \sqrt{0.0204^2 + 0.022^2} = 0.03$

Value of the test statistic:

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

$z = \frac{0.08 - 0}{0.03}$

$z = 2.67$

Question a:

P-value of the test and decision:

The p-value of the test is the probability that the sample proportion differs from 0 by at least 0.08, which is P(|Z| > 2.670, which is 2 multiplied by the p-value of Z = -2.67.

Looking at the z-table, Z = -2.67 has a p-value of 0.0038.

2*0.0038 = 0.0076

The p-value of the test is 0.0076 < 0.05, which means that there is evidence of a difference between males and females in the proportion who said they prefer window tinting as a luxury upgrade at the 0.05.