College

Nationwide, 76% of first-year college students live on campus. The director of housing at a large private institution would like to know if it is significantly different at his school. He selects a random sample of 46 first-year students and finds that 78% of them live on campus. Do these data provide convincing evidence that the true proportion of all first-year students who attend this private institution and live on campus differs from the national average? Use [tex]$\alpha=0.05$[/tex].

Which of the following statements are true? Check all that apply.

- [tex]$H_0: p=0.76$[/tex]
- [tex]$H_0: p=0.89$[/tex]
- The random condition is met.
- The 10% condition is met.
- The large counts condition is met.
- The test is a z-test for one proportion.

Answer :

To determine whether the proportion of first-year students living on campus at the private institution differs significantly from the national average, we can perform a hypothesis test concerning a single proportion. Here is how we can go about it:

### Step 1: Define the Hypotheses

- Null Hypothesis ([tex]\(H_0\)[/tex]): The proportion of first-year students at the private institution who live on campus is equal to the national average, [tex]\(p = 0.76\)[/tex].
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): The proportion of first-year students at the private institution who live on campus is different from the national average, [tex]\(p \neq 0.76\)[/tex].

### Step 2: Check the Conditions for Performing a Z-Test for One Proportion

1. Random Condition: The sample should be a simple random sample from the population. We assume this condition is met based on the problem stating it is a random sample.

2. 10% Condition: The sample size should be less than 10% of the population. Since the sample consists of 46 students, this condition is considered met if the total number of first-year students at the institution is more than 460.

3. Large Counts Condition: For the normal approximation to be suitable, both [tex]\(np\)[/tex] and [tex]\(n(1-p)\)[/tex] need to be at least 10:
- [tex]\(n = 46\)[/tex], [tex]\(p = 0.76\)[/tex]
- Calculate [tex]\(np = 46 \times 0.76 = 34.96\)[/tex]
- Calculate [tex]\(n(1-p) = 46 \times (1-0.76) = 11.04\)[/tex]
- Both calculations are greater than 10, so this condition is met.

### Step 3: Conduct the Test

Since all conditions are met, a z-test for one proportion is appropriate for this analysis.

- The sample proportion found was 78%, or 0.78.
- The question specifically focuses on whether the sample proportion of 0.78 is significantly different from the national proportion of 0.76.

### Conclusion

- [tex]\(H_0: p=0.76\)[/tex] is true and is the correct null hypothesis.
- [tex]\(H_0: p=0.89\)[/tex] is false and not related to this test.

Finally, the statements:
- "The random condition is met.",
- "The 10% condition is met.",
- "The large counts condition is met.",
- "The test is a z-test for one proportion."

Are all true, confirming that the hypothesis test set-up is appropriate given these conditions.