In a random sample of 500 students, 80% say they have a part-time job. An approximate 95% confidence
interval for the proportion of students with part-time jobs is:
(1) (0.758; 0.842)
(2) (0.765; 0.835)
(3) (0.770; 0.830)
(4) (0.800; 0.835)
(5) (0.782; 0.819)
To calculate the approximate 95% confidence interval for the proportion of students with part-time jobs, we can use the formula:
CI = p̂ ± z * √(p̂(1-p̂)/n)
where:
- p̂ is the sample proportion (80% or 0.8 in this case),
- z is the z-score corresponding to the desired confidence level (95% in this case, which corresponds to a z-score of approximately 1.96),
- n is the sample size (500 in this case).
Plugging in the values, we can calculate the confidence interval as follows:
CI = 0.8 ± 1.96 * √(0.8(1-0.8)/500)
CI = 0.8 ± 1.96 * √(0.16/500)
CI = 0.8 ± 1.96 * √(0.00032)
CI = 0.8 ± 1.96 * 0.01788854382
CI = 0.8 ± 0.03509196999
CI ≈ (0.765; 0.835)
Therefore, the approximate 95% confidence interval for the proportion of students with part-time jobs is option (2) (0.765; 0.835).