Final answer:
To construct a 95% confidence interval for the proportion of juniors with a passport, calculate the sample proportion and the standard error, then apply the 1.96 z-score for a 95% confidence level. The confidence interval is (0.245, 0.352), rounded to the nearest thousandth.
Step-by-step explanation:
We need to construct a 95% confidence interval for the proportion of junior college students that have a passport based on the sample data: 335 students surveyed, 100 of which said they have a passport.
Firstly, calculate the sample proportion (p) by dividing the number of students with a passport by the total number of students surveyed:
p = 100 / 335 = 0.2985 (rounded to four decimal places)
Next, use the standard error for the proportion, which is given by the formula SE = sqrt[p(1-p)/n] where n is the sample size:
SE = sqrt[0.2985 * (1 - 0.2985) / 335] ≈ 0.0274 (rounded to four decimal places)
For a 95% confidence interval, we use the z-score associated with a 95% confidence level. This is approximately 1.96 for a two-tailed test.
CI = p ± z * SE
CI = 0.2985 ± 1.96 * 0.0274
CI = 0.2985 ± 0.0537
Lower limit = 0.2985 - 0.0537 = 0.2448
Upper limit = 0.2985 + 0.0537 = 0.3522
Therefore, the 95% confidence interval for the true proportion of junior college students that have a passport rounds to (0.245, 0.352) when rounded to the nearest thousandth.