A simple random sample is conducted of 1486 young adults who were once college students seeking bachelor’s degrees at a specific university. The study found that 780 of these young adults actually earned bachelor’s degrees. Use a 0.05 significance level to test the claim that a majority of the students at this university earn bachelor’s degrees. State the following
Null and alternative hypotheses:
Left -, right- or two tail test:
Test statistic
Critical value (s) (and draw the rejection region)
State rejection region:
P-value (draw this area, too)
Decision to reject/ fail to reject the null hypothesis (based on both tests)
Fully state your conclusion (in the context of the problem)
Of 99 adults selected randomly from one town, 63 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance. Show all work. (Round confidence interval limits to three decimal positions)
Confidence: ___________<p<________________
For the first question:
Null hypothesis (H0): Majority of the students earn bachelor's degrees at this university (p >= 0.5)
Alternative hypothesis (Ha): Majority of the students do not earn bachelor's degrees at this university (p < 0.5)
This is a one-tailed test because we are testing whether the proportion is significantly less than 0.5.
Test statistic: We will use the z-test statistic since the sample size is large (n > 30) and we know the population proportion.
z = (p^ - p) / sqrt (p * (1-p) / n)
Where p^ is the sample proportion (780/1486), p is the hypothesized proportion (0.5), and n is the sample size.
Critical value: For a significance level of 0.05, the critical value is -1.645 (since it is a left-tailed test). The rejection region is any z-score less than or equal to -1.645.
p-value: We can use the test statistic to find the p-value associated with the observed value. The p-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true.
Decision: If the test statistic falls in the rejection region or the p-value is less than 0.05, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Conclusion: Based on the results of the test, we either conclude that there is sufficient evidence to support the claim that a majority of the students at this university earn bachelor's degrees, or we fail to reject that claim.
For the second question:
To find a confidence interval, we can use the formula:
CI = p^ ± z * sqrt(p^ * (1-p^) / n)
Where p^ is the sample proportion (63/99), z is the z-score associated with the desired confidence level (90% corresponds to a z-score of 1.645 for a two-tailed test), and n is the sample size.
Confidence interval: Plugging in the values, we have:
CI = 0.636 ± 1.645 * sqrt(0.636 * (1-0.636) / 99)
Round the confidence interval limits to three decimal places.
So, the confidence interval for the true proportion of all adults in the town who have health insurance is:
0.571 < p < 0.701
For the first problem:
Null hypothesis: The proportion of students at this university who earn bachelor's degrees is equal to or less than 50%.
Alternative hypothesis: The proportion of students at this university who earn bachelor's degrees is greater than 50%.
This is a one-tailed test since we are testing if there is a majority (more than 50%).
Test statistic: In this case, we will use a z-test. The formula for the test statistic is:
z = (p - P) / √(P(1-P) / n)
where p is the sample proportion, P is the hypothesized proportion (in this case, 0.5), and n is the sample size.
Critical value: Since the significance level is 0.05 (confidence level of 95%), we will use zα = 1.645 to compare with the test statistic. The rejection region is in the upper tail, z ≥ 1.645.
P-value: The P-value is the probability of obtaining a test statistic equal to or more extreme than the observed value, assuming the null hypothesis is true.
Decision: If the test statistic falls in the rejection region or the P-value is less than the significance level (0.05), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Conclusion: Based on the test results, we will either reject or fail to reject the null hypothesis, and provide a conclusion regarding whether a majority of the students at this university earn bachelor's degrees.
For the second problem:
Given information:
Sample size (n) = 99
Number of adults with health insurance (x) = 63
Confidence interval formula:
p ± z * √(p(1-p) / n)
where p is the sample proportion, z is the z-score corresponding to the desired confidence level, and n is the sample size.
Calculating the sample proportion:
p = x / n = 63 / 99 = 0.6364 (rounded to four decimal places)
Confidence interval:
Using a 90% confidence level (alpha = 0.1) results in a z-score of 1.645.
Plugging in the values:
Confidence interval = 0.6364 ± 1.645 * √(0.6364(1-0.6364) / 99)
Calculating the confidence interval:
lower limit = 0.6364 - 1.645 * √(0.6364(1-0.6364) / 99)
upper limit = 0.6364 + 1.645 * √(0.6364(1-0.6364) / 99)
Rounding to three decimal places, the confidence interval is:
Confidence: 0.562 < p < 0.711