An SRS of 40 San Diego County Schools graduates showed that 26 of the 40 enrolled in a college or university right out of high school.

a. Find a 90% confidence interval for the proportion of all San Diego County Schools graduates who enrolled in a college or university right out of high school. Be sure to justify the formula (assumptions / conditions).

b. The proportion for all graduates across the nation who enroll in a college or university right out of high school is 52%. Is there enough evidence at the 5% significance level to suggest that the proportion of San Diego County graduates is greater than the national average? Be sure to include all the parts of the test.

a. To find a 90% confidence interval for the proportion of all San Diego County Schools graduates who enrolled in a college or university right out of high school, we can use the formula for confidence intervals for proportions.

The formula for the confidence interval for a proportion is:

CI = p̂ ± Z * sqrt((p̂ * (1 - p̂))/n)

where:
- CI is the confidence interval
- p̂ is the sample proportion (number of successes divided by sample size)
- Z is the z-score corresponding to the desired confidence level
- sqrt represents the square root function
- n is the sample size

In this case, we have:
- p̂ = 26/40 = 0.65
- n = 40
- The desired confidence level is 90%, which corresponds to a z-score of 1.645 (obtained from a z-table).

Plugging these values into the formula:

CI = 0.65 ± 1.645 * sqrt((0.65 * (1 - 0.65))/40)

Calculating this expression:

CI = 0.65 ± 1.645 * sqrt(0.2275/40)

CI = 0.65 ± 1.645 * 0.0694

CI ≈ 0.65 ± 0.1141

Therefore, the 90% confidence interval for the proportion of all San Diego County Schools graduates who enrolled in a college or university right out of high school is approximately (0.536, 0.764).

b. To determine if there is enough evidence at the 5% significance level to suggest that the proportion of San Diego County graduates who enroll in a college or university right out of high school is greater than the national average, we can perform a hypothesis test.

Let's set up the hypotheses:

Null Hypothesis (H0): The proportion of San Diego County graduates who enroll in a college or university right out of high school is equal to the national average (p = 0.52).
Alternative Hypothesis (Ha): The proportion of San Diego County graduates who enroll in a college or university right out of high school is greater than the national average (p > 0.52).

To test these hypotheses, we can use a one-sample z-test.

The formula for the z-statistic is:

z = (p̂ - p0) / sqrt((p0 * (1 - p0))/n)

where:
- z is the test statistic
- p̂ is the sample proportion (number of successes divided by sample size)
- p0 is the hypothesized population proportion under the null hypothesis
- sqrt represents the square root function
- n is the sample size

In this case, we have:
- p̂ = 0.65 (from the sample)
- p0 = 0.52 (national average)
- n = 40

Plugging these values into the formula:

z = (0.65 - 0.52) / sqrt((0.52 * (1 - 0.52))/40)

Calculating this expression:

z = 0.13 / sqrt(0.2496/40)

z ≈ 0.13 / 0.0789

z ≈ 1.647

The critical value for a one-tailed test at the 5% significance level is approximately 1.645 (obtained from a z-table). Since our test statistic (z = 1.647) is greater than the critical value, we reject the null hypothesis.

Therefore, there is enough evidence at the 5% significance level to suggest that the proportion of San Diego County graduates who enroll in a college or university right out of high school is greater than the national average.

a. To find a 90% confidence interval for the proportion of all San Diego County Schools graduates who enrolled in a college or university right out of high school, we can use the formula:

CI = p̂ ± Z * √( (p̂ * (1 - p̂)) / n )

Where:
CI = Confidence Interval
p̂ = Sample proportion (26 / 40 = 0.65)
Z = Z-score corresponding to the desired confidence level (90% confidence corresponds to a Z-score of 1.645)
n = Sample size (40)

Plugging in the values, we get:

CI = 0.65 ± 1.645 * √( (0.65 * (1 - 0.65)) / 40 )

Calculating the values inside the square root:

CI = 0.65 ± 1.645 * √(0.22775 / 40)

CI = 0.65 ± 1.645 * √(0.00569375)

CI = 0.65 ± 1.645 * 0.07545

CI = 0.65 ± 0.124

Therefore, the 90% confidence interval for the proportion of all San Diego County Schools graduates who enrolled in a college or university right out of high school is (0.526, 0.774).

In terms of assumptions and conditions, for the formula to be valid, we need to assume that the sample is representative of the population, the observations are independent, and the sample size is large enough for the Central Limit Theorem to apply (which is satisfied with a sample size of 40).

b. To test whether the proportion of San Diego County graduates is greater than the national average (52%), we can perform a hypothesis test using the Z-test for comparing two population proportions.

Null hypothesis (H0): The proportion of San Diego County graduates is equal to the national average (p = 0.52).
Alternative hypothesis (Ha): The proportion of San Diego County graduates is greater than the national average (p > 0.52).

Test statistic formula:

Z = (p̂ - p) / √( (p * (1 - p)) / n )

Where:
Z = Test statistic
p̂ = Sample proportion (0.65)
p = Population proportion (0.52)
n = Sample size (40)

Plugging in the values, we get:

Z = (0.65 - 0.52) / √( (0.52 * (1 - 0.52)) / 40 )

Calculating inside the square root:

Z = 0.13 / √(0.2496 / 40)

Z = 0.13 / √(0.00624)

Z = 0.13 / 0.079

Z = 1.646

The test statistic (Z) is 1.646.

Next, we need to find the p-value associated with this test statistic. Assuming this is a one-tailed test (Ha: p > 0.52), we can look up the p-value corresponding to 1.646 in a Z-table or use statistical software. Let's assume the p-value is 0.05.

Since the p-value (0.05) is less than the significance level (0.05), we reject the null hypothesis. There is enough evidence to suggest that the proportion of San Diego County graduates is greater than the national average.

In conclusion, based on the hypothesis test, there is enough evidence to suggest that the proportion of San Diego County graduates who enroll in a college or university right out of high school is greater than the national average.

I'll get you started.

CI90 = p ± (1.645)[√(pq/n)]
...where p = x/n, q = 1 - p, and n = sample size.
Note: ± 1.645 represents 90% confidence interval.

For p in your problem: 26/40
For q in your problem: 14/40
n = 40

I let you take it from here to calculate the interval. (Note: convert all fractions to decimals.)

For b):
You can try a proportional one-sample z-test for this one since this problem is using proportions.

Using a formula for a proportional one-sample z-test with your data included, we have:
z = .65 - .52 -->test value (26/40) is .65) minus population value (.52) divided by
√[(.52)(.48)/40] --> .48 represents 1-.52 and 40 is sample size.

Finish the calculation.

Once you have the z-test statistic calculated above, compare the test statistic to the .05 significance level for a one-tailed test. If the test statistic exceeds the critical value from the table at the .05 level for a one-tailed test, then reject the null and conclude the proportion is greater than the national average. If the test statistic does not exceed the critical value from the table, do not reject the null. You cannot conclude a difference in this case.

I hope this will help.