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
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
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
and standard deviation
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](https://tex.z-dn.net/?f=p_A%20%3D%20%5Cfrac%7B38%7D%7B78%7D%20%3D%200.4872)
![s_A = \sqrt{\frac{0.4872*0.5128}{78}} = 0.0566](https://tex.z-dn.net/?f=s_A%20%3D%20%5Csqrt%7B%5Cfrac%7B0.4872%2A0.5128%7D%7B78%7D%7D%20%3D%200.0566)
Home:
46 out of 84, so:
![p_H = \frac{46}{84} = 0.5476](https://tex.z-dn.net/?f=p_H%20%3D%20%5Cfrac%7B46%7D%7B84%7D%20%3D%200.5476)
![s_H = \sqrt{\frac{0.5476*0.4524}{84}} = 0.0543](https://tex.z-dn.net/?f=s_H%20%3D%20%5Csqrt%7B%5Cfrac%7B0.5476%2A0.4524%7D%7B84%7D%7D%20%3D%200.0543)
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](https://tex.z-dn.net/?f=H_0%3A%20p_A%20-%20p_H%20%3D%200)
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](https://tex.z-dn.net/?f=H_1%3A%20p_A%20-%20p_H%20%5Cneq%200)
The test statistic is:
In which X is the sample mean,
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](https://tex.z-dn.net/?f=%5Cmu%20%3D%200)
From the samples:
![X = p_A - p_H = 0.4872 - 0.5476 = -0.0604](https://tex.z-dn.net/?f=X%20%3D%20p_A%20-%20p_H%20%3D%200.4872%20-%200.5476%20%3D%20-0.0604)
![s = \sqrt{s_A^2 + s_H^2} = \sqrt{0.0566^2 + 0.0543^2} = 0.0784](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7Bs_A%5E2%20%2B%20s_H%5E2%7D%20%3D%20%5Csqrt%7B0.0566%5E2%20%2B%200.0543%5E2%7D%20%3D%200.0784)
Value of the test statistic:
![z = -0.77](https://tex.z-dn.net/?f=z%20%3D%20-0.77)
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.