Question

A student at a four-year college claims that mean enrollment at two-year colleges is lower than at four-year colleges in the United States. Two surveys are conducted. Of the 35 two-year colleges surveyed, the mean enrollment was 5,068 with a standard deviation of 4,777. Of the 35 four-year colleges surveyed, the mean enrollment was 5,466 with a standard deviation of 8,191. Conduct a hypothesis test at the 5% significance level.

Answer 1

Answer:

The p-value of the test is of 0.0708 > 0.05, which means that there is not enough evidence for the claim the the mean enrollment at two-year colleges is lower than at four-year colleges in the United States.

Step-by-step explanation:

To solve this question, we need to understand subtraction of normal variables and the central limit theorem.

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.

35 two-year colleges surveyed, the mean enrollment was 5,068 with a standard deviation of 4,777.

This means that $\mu_2 = 5068, s_2 = \frac{4777}{\sqrt{35}}$

Of the 35 four-year colleges surveyed, the mean enrollment was 5,466 with a standard deviation of 8,191.

This means that $\mu_4 = 5466, s_4 = \frac{8191}{\sqrt{35}}$

A student at a four-year college claims that mean enrollment at two-year colleges is lower than at four-year colleges in the United States.

At the null hypothesis, we test if is at least equal, that is, the subtraction of the mean enrollment at 2 years colleges subtracted by the mean enrollment at 4 years colleges is at least 0. So

$H_0: \mu_2 - \mu_4 \geq 0$

At the alternative hypothesis, we test if is less, that is, the subtraction is less than 0.

$H_1: \mu_2 - \mu_4 < 0$

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:

$X = \mu_2 - \mu_4 = 5068 - 5466 = -398$

$s = \sqrt{s_2^2+s_4^2} =\sqrt{(\frac{4777}{\sqrt{35}})^2+(\frac{8191}{\sqrt{35}})^2} = 270.92$

Test statistic:

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

$z = \frac{-398 - 0}{270.92}$

$z = -1.47$

P-value of the test:

The p-value of the test is the probability of a difference of 398 or more, which is the p-value of z = -1.47.

Looking at the z-table, z = -1.47 has a p-value of 0.0708.

The p-value of the test is of 0.0708 > 0.05, which means that there is not enough evidence for the claim the the mean enrollment at two-year colleges is lower than at four-year colleges in the United States.