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:
Home:
46 out of 84, so:
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:
At the alternative hypothesis, we test if there is a difference, that is, the subtraction of the proportions is different of 0, so:
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
From the samples:
Value of the test statistic:
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.