Mathematics, 28.05.2021 14:00, BluVoid9255

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 ofchoice, 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 alternativehypotheses, 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.

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Answers

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.

answered: Guest

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answered: Guest

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step-by-step explanation: