To test the student's claim that the mean enrollment at four-year colleges is higher than at two-year colleges, we will perform a hypothesis test for the difference between two population means with independent samples.

Preliminary Analysis

1. It is safe to assume that we are comparing populations of all students enrolled at 4-year and 2-year colleges.
2. We assume the sample data is representative of the populations.

Now, let's proceed to the hypothesis test:

Hypothesis Statements

• Null Hypothesis ( $H_0$ ): $\mu_1 \leq \mu_2$ (the mean enrollment at four-year colleges is less than or equal to that at two-year colleges)
• Alternative Hypothesis ( $H_1$ ): $\mu_1 > \mu_2$ (the mean enrollment at four-year colleges is greater than that at two-year colleges)

Step-by-Step Solution

Given Data:

• Sample mean of four-year colleges ($\bar{X}_1$) = 6067

• Sample standard deviation of four-year colleges ($S_1$) = 772

• Sample size of four-year colleges ($n_1$) = 35

• Sample mean of two-year colleges ($\bar{X}_2$) = 5985

• Sample standard deviation of two-year colleges ($S_2$) = 712

• Sample size of two-year colleges ($n_2$) = 35

Significance Level:

• $\alpha = 0.1$

Step 1: Calculate the Test Statistic

We use the formula for the test statistic for the difference in means:

$z = \frac{(\bar{X}_1 - \bar{X}_2)}{\sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}}$

Plugging in the values:

$z = \frac{(6067 - 5985)}{\sqrt{\frac{772^2}{35} + \frac{712^2}{35}}}$

Calculate each component:

1. $\bar{X}_1 - \bar{X}_2 = 6067 - 5985 = 82$
2. $\frac{772^2}{35} \approx 17024.5143$
3. $\frac{712^2}{35} \approx 14491.5429$
4. Sum these variances: $17024.5143 + 14491.5429 = 31516.0571$
5. Take the square root: $\sqrt{31516.0571} \approx 177.55$

Then,

$z = \frac{82}{177.55} \approx 0.4618$

Step 2: Find the p-value

Since this is a one-tailed test, we find the p-value for $z = 0.4618$.

Using the standard normal distribution table, the p-value associated with $z = 0.4618$ is approximately $0.3222$.

Step 3: Make a Decision

• Since the p-value $0.3222$ is greater than the significance level $\alpha = 0.1$, we fail to reject the null hypothesis.

Conclusion

There is not sufficient evidence to support the claim that enrollment is higher at four-year colleges than at two-year colleges.

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Would you like a more detailed breakdown of any step?

Here are 5 related questions for further practice:

1. How would the conclusion change if the significance level were set at 0.05 instead of 0.1?
2. If the sample sizes were larger, say 100 colleges for each type, how would this affect the test statistic?
3. What if we wanted to test if the enrollment is different (not specifically higher or lower)? How would the hypotheses and test change?
4. How would using a t-test affect the calculation of the test statistic and interpretation?
5. What is the power of this test, and how could we increase it?

Tip: When calculating p-values, remember to account for whether the test is one-tailed or two-tailed, as this affects the rejection region.