Question

People were asked if they owned an artificial Christmas tree. Of 78 people who lived in an apartment, 38 own an artificial Christmas tree. Also it was learned that of 84 people who own their home, 46 own an artificial Christmas tree. Is there a significant difference in the proportion of apartment dwellers and home owners who own artificial Christmas trees

Answer 1

Answer:

The p-value of the test is 0.4414, higher than the standard significance level of 0.05, which means that there is not a a significant difference in the proportion of apartment dwellers and home owners who own artificial Christmas trees.

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.

Apartment:

38 out of 78, so:

$p_A = \frac{38}{78} = 0.4872$

$s_A = \sqrt{\frac{0.4872*0.5128}{78}} = 0.0566$

Home:

46 out of 84, so:

$p_H = \frac{46}{84} = 0.5476$

$s_H = \sqrt{\frac{0.5476*0.4524}{84}} = 0.0543$

Test if the there a significant difference in the proportion of apartment dwellers and home owners who own artificial Christmas trees:

At the null hypothesis, we test if there is no difference, that is, the subtraction of the proportions is equal to 0, so:

$H_0: p_A - p_H = 0$

At the alternative hypothesis, we test if there is a difference, that is, the subtraction of the proportions is different of 0, so:

$H_1: p_A - p_H \neq 0$

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 samples:

$X = p_A - p_H = 0.4872 - 0.5476 = -0.0604$

$s = \sqrt{s_A^2 + s_H^2} = \sqrt{0.0566^2 + 0.0543^2} = 0.0784$

Value of the test statistic:

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

$z = \frac{-0.0604 - 0}{0.0784}$

$z = -0.77$

P-value of the test and decision:

The p-value of the test is the probability of the difference being of at least 0.0604, to either side, plus or minus, which is P(|z| > 0.77), given by 2 multiplied by the p-value of z = -0.77.

Looking at the z-table, z = -0.77 has a p-value of 0.2207.

2*0.2207 = 0.4414

The p-value of the test is 0.4414, higher than the standard significance level of 0.05, which means that there is not a a significant difference in the proportion of apartment dwellers and home owners who own artificial Christmas trees.