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.
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.
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