Answer :
To determine if the proportion of first-year students living on campus at a large private institution significantly differs from the national average of 76%, we will perform a hypothesis test using a significance level of [tex]\(\alpha = 0.05\)[/tex]. Here’s the step-by-step solution:
### Step 1: State the Hypotheses
- Null hypothesis ([tex]\(H_0\)[/tex]): [tex]\(p = 0.76\)[/tex], where [tex]\(p\)[/tex] is the true proportion of all first-year students at the private institution who live on campus.
- Alternative hypothesis ([tex]\(H_A\)[/tex]): [tex]\(p \neq 0.76\)[/tex].
### Step 2: Check the Conditions for a One-Proportion Z-Test
1. Random Condition: The sample is assumed to be randomly selected.
2. 10% Condition: The sample size should be less than 10% of the population. Since we are considering students at a large institution, 46 students are likely to be less than 10% of the student population.
3. Large Counts Condition: Both [tex]\(n \times p_0\)[/tex] and [tex]\(n \times (1 - p_0)\)[/tex] should be at least 10.
[tex]\[
46 \times 0.76 = 34.96 \quad \text{and} \quad 46 \times (1 - 0.76) = 11.04
\][/tex]
Both values are greater than 10, so this condition is met.
### Step 3: Calculate the Test Statistic
We use the sample proportion ([tex]\(\hat{p} = 0.78\)[/tex]) to calculate the Z-score:
- Population proportion ([tex]\(p_0 = 0.76\)[/tex])
- Sample size ([tex]\(n = 46\)[/tex])
First, we calculate the standard error (SE):
[tex]\[
\text{SE} = \sqrt{\frac{p_0 (1 - p_0)}{n}} = \sqrt{\frac{0.76 \times 0.24}{46}} \approx 0.06297
\][/tex]
Next, we calculate the Z-score:
[tex]\[
Z = \frac{\hat{p} - p_0}{\text{SE}} = \frac{0.78 - 0.76}{0.06297} \approx 0.3176
\][/tex]
### Step 4: Calculate the P-Value
The P-value is computed using the standard normal distribution. Since this is a two-tailed test, we calculate the area in both tails.
[tex]\[
P(\text{value}) = 2 \times (1 - \Phi(|0.3176|)) \approx 0.7508
\][/tex]
### Step 5: Conclusion
Compare the P-value with the significance level [tex]\(\alpha = 0.05\)[/tex]:
[tex]\[
0.7508 > 0.05
\][/tex]
Since the P-value is greater than [tex]\(\alpha\)[/tex], we fail to reject the null hypothesis.
### Summary of True Statements:
- [tex]\(H_0: p = 0.76\)[/tex]
- The random condition is met.
- The 10% condition is met.
- The large counts condition is met.
- The test is a [tex]\(z\)[/tex]-test for one proportion.
Thus, the data does not provide convincing evidence that the true proportion of all first-year students at this private institution who live on campus differs from the national average of 76%.
### Step 1: State the Hypotheses
- Null hypothesis ([tex]\(H_0\)[/tex]): [tex]\(p = 0.76\)[/tex], where [tex]\(p\)[/tex] is the true proportion of all first-year students at the private institution who live on campus.
- Alternative hypothesis ([tex]\(H_A\)[/tex]): [tex]\(p \neq 0.76\)[/tex].
### Step 2: Check the Conditions for a One-Proportion Z-Test
1. Random Condition: The sample is assumed to be randomly selected.
2. 10% Condition: The sample size should be less than 10% of the population. Since we are considering students at a large institution, 46 students are likely to be less than 10% of the student population.
3. Large Counts Condition: Both [tex]\(n \times p_0\)[/tex] and [tex]\(n \times (1 - p_0)\)[/tex] should be at least 10.
[tex]\[
46 \times 0.76 = 34.96 \quad \text{and} \quad 46 \times (1 - 0.76) = 11.04
\][/tex]
Both values are greater than 10, so this condition is met.
### Step 3: Calculate the Test Statistic
We use the sample proportion ([tex]\(\hat{p} = 0.78\)[/tex]) to calculate the Z-score:
- Population proportion ([tex]\(p_0 = 0.76\)[/tex])
- Sample size ([tex]\(n = 46\)[/tex])
First, we calculate the standard error (SE):
[tex]\[
\text{SE} = \sqrt{\frac{p_0 (1 - p_0)}{n}} = \sqrt{\frac{0.76 \times 0.24}{46}} \approx 0.06297
\][/tex]
Next, we calculate the Z-score:
[tex]\[
Z = \frac{\hat{p} - p_0}{\text{SE}} = \frac{0.78 - 0.76}{0.06297} \approx 0.3176
\][/tex]
### Step 4: Calculate the P-Value
The P-value is computed using the standard normal distribution. Since this is a two-tailed test, we calculate the area in both tails.
[tex]\[
P(\text{value}) = 2 \times (1 - \Phi(|0.3176|)) \approx 0.7508
\][/tex]
### Step 5: Conclusion
Compare the P-value with the significance level [tex]\(\alpha = 0.05\)[/tex]:
[tex]\[
0.7508 > 0.05
\][/tex]
Since the P-value is greater than [tex]\(\alpha\)[/tex], we fail to reject the null hypothesis.
### Summary of True Statements:
- [tex]\(H_0: p = 0.76\)[/tex]
- The random condition is met.
- The 10% condition is met.
- The large counts condition is met.
- The test is a [tex]\(z\)[/tex]-test for one proportion.
Thus, the data does not provide convincing evidence that the true proportion of all first-year students at this private institution who live on campus differs from the national average of 76%.
