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

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.

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 = (\sigma)/(√(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{(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 = (4777)/(√(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 = (8191)/(√(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 = (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 = √(s_2^2+s_4^2) =\sqrt{((4777)/(√(35)))^2+((8191)/(√(35)))^2} = 270.92`

Test statistic:

`z = (X - \mu)/(s)`

`z = (-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.