Confidence interval with one-proportion z-test
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Scenario: You are curious the how many freshmen students from Crammer Nation University joined Greek Life this year. You conduct a random sample of 50 randomly selected freshmen and find that 18 of them joined a Greek Life chapter. What is the 95% confidence interval?
| Clue | Insight |
|---|---|
| We're given a sample proportion of 18 out of 50. | We're working with a proportion here. |
| We've gotta find the 95% confidence interval for the true population proportion. | We're working with confidence intervals. |
| Assumption | Validate |
|---|---|
| Sample is randomly selected from the population. | We see "randomly selected" in the scenario. ✅ |
| The sample size (n) is less than or equal to 10% of the population size (N). | We can assume a sample of 50 students is less than 10% of the entire student population at Crammer Nation University. ✅ |
There are 10 successes and 10 failures in the sample OR np >= 10 and nq >= 10. | We've got 18 [students who] joined a Greek Life, leaving 50 - 18 = 32 others who didn't! Both those are greater than 10! ✅ |
p-hat is the sample proportion we've observed.
q-hat is the sample proportion of failure we've observed (a.k.a. the opposite of p-hat).
Z* is our z-score critical value.
n is our sample size.
p-hat = 18 / 50 = 0.36
q-hat = 1 - 0.36 = 0.64
n = 50
Z* = 1.960
| Confidence Level | Z* value |
|---|---|
| 90% | 1.645 |
| 95% | 1.960 |
| 99% | 2.576 |
CI = 0.36 ± 1.960 √[(0.36 * 0.64) / 50]
CI = 0.36 ± 1.960 √[(0.2304) / 50]
CI = 0.36 ± 1.960 √[0.0046]
CI = 0.36 ± 1.960 (0.0678)
CI = 0.36 ± 0.1329
CI = (0.2271, 0.4929)
Answer: We are 95% confident that the true population proportion of Crammer Nation University freshmen who joined Greek Life this year is between 0.2271 and 0.4929.