Answer :
To determine if there is convincing evidence that the proportion of first-year students living on campus at the private institution is different from the national average of 76%, we can conduct a hypothesis test for one proportion using a significance level of [tex]\(\alpha = 0.05\)[/tex].
### Step-by-step Solution:
1. State the Null and Alternative Hypotheses:
- Null Hypothesis ([tex]\(H_0\)[/tex]): [tex]\(p = 0.76\)[/tex]
(The proportion of first-year students living on campus is the same as the national average)
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): [tex]\(p \neq 0.76\)[/tex]
(The proportion of first-year students living on campus is different from the national average)
2. Check the Conditions for a Hypothesis Test:
- Random Condition: The sample is a random sample of 46 first-year students.
- 10% Condition: The sample size of 46 is less than 10% of all first-year students at the institution, assuming the institution has more than 460 first-year students.
- Large Counts Condition: To use a normal approximation, we need to check if [tex]\(np\)[/tex] and [tex]\(n(1-p)\)[/tex] are both greater than 10.
- [tex]\(np = 46 \times 0.76 = 34.96\)[/tex]
- [tex]\(n(1-p) = 46 \times 0.24 = 11.04\)[/tex]
Both are greater than 10, so this condition is met.
3. Calculate the Standard Error:
[tex]\[
\text{Standard Error} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.76 \times 0.24}{46}} \approx 0.06297
\][/tex]
4. Calculate the z-score:
The formula for the z-score in this test is:
[tex]\[
z = \frac{\text{sample proportion} - \text{null proportion}}{\text{Standard Error}} = \frac{0.78 - 0.76}{0.06297} \approx 0.318
\][/tex]
5. Determine the p-value:
Since we are conducting a two-tailed test (because [tex]\(H_a: p \neq 0.76\)[/tex]), we find the p-value by finding the probability of observing a z-score as extreme as the calculated one in either tail of the normal distribution:
[tex]\[
\text{p-value} = 2 \times (1 - \text{CDF}(|z|)) \approx 0.751
\][/tex]
6. Make a Decision:
- Compare the p-value to the significance level [tex]\(\alpha = 0.05\)[/tex].
- Since [tex]\(0.751 > 0.05\)[/tex], we fail to reject the null hypothesis.
7. Conclusion:
There is not enough evidence to conclude that the proportion of first-year students living on campus at this institution is different from the national average of 76%. Therefore, the data does not provide convincing evidence to suggest a deviation from the national average.
### Step-by-step Solution:
1. State the Null and Alternative Hypotheses:
- Null Hypothesis ([tex]\(H_0\)[/tex]): [tex]\(p = 0.76\)[/tex]
(The proportion of first-year students living on campus is the same as the national average)
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): [tex]\(p \neq 0.76\)[/tex]
(The proportion of first-year students living on campus is different from the national average)
2. Check the Conditions for a Hypothesis Test:
- Random Condition: The sample is a random sample of 46 first-year students.
- 10% Condition: The sample size of 46 is less than 10% of all first-year students at the institution, assuming the institution has more than 460 first-year students.
- Large Counts Condition: To use a normal approximation, we need to check if [tex]\(np\)[/tex] and [tex]\(n(1-p)\)[/tex] are both greater than 10.
- [tex]\(np = 46 \times 0.76 = 34.96\)[/tex]
- [tex]\(n(1-p) = 46 \times 0.24 = 11.04\)[/tex]
Both are greater than 10, so this condition is met.
3. Calculate the Standard Error:
[tex]\[
\text{Standard Error} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.76 \times 0.24}{46}} \approx 0.06297
\][/tex]
4. Calculate the z-score:
The formula for the z-score in this test is:
[tex]\[
z = \frac{\text{sample proportion} - \text{null proportion}}{\text{Standard Error}} = \frac{0.78 - 0.76}{0.06297} \approx 0.318
\][/tex]
5. Determine the p-value:
Since we are conducting a two-tailed test (because [tex]\(H_a: p \neq 0.76\)[/tex]), we find the p-value by finding the probability of observing a z-score as extreme as the calculated one in either tail of the normal distribution:
[tex]\[
\text{p-value} = 2 \times (1 - \text{CDF}(|z|)) \approx 0.751
\][/tex]
6. Make a Decision:
- Compare the p-value to the significance level [tex]\(\alpha = 0.05\)[/tex].
- Since [tex]\(0.751 > 0.05\)[/tex], we fail to reject the null hypothesis.
7. Conclusion:
There is not enough evidence to conclude that the proportion of first-year students living on campus at this institution is different from the national average of 76%. Therefore, the data does not provide convincing evidence to suggest a deviation from the national average.
