In a certain city, there are 10,000 persons age 18 to 24. A simple random sample of
500 such persons is drawn, of whom 200 turn out to be currently enrolled in college. Find a 90%
con¯dence interval for the percentage of persons age 18 to 24 in the city who are attending college.
Try using a binomial proportion CI formula:
CI90 = p + or - 1.645(√pq/n)
...where p = proportion in the problem, q = 1 - p, n = sample size, √ = square root, and + or - 1.645 represents 90% CI using a z-table.
Your data:
p = 200/500
q = 1 - p
n = 500
Convert all fractions to decimals. Substitute values into the formula and determine the confidence interval.
I hope this will help get you started.
500.12
To find a 90% confidence interval for the percentage of persons age 18 to 24 in the city who are attending college, you can use the following formula:
Confidence Interval = Sample Proportion ± Margin of Error
To calculate the sample proportion, divide the number of individuals in the sample who are currently enrolled in college (200) by the sample size (500):
Sample Proportion = (Number of individuals currently enrolled in college) / (Sample size)
Sample Proportion = 200 / 500 = 0.4
The margin of error depends on the sample size and the desired confidence level. For a 90% confidence level, you need to find the critical value of the standard normal distribution (Z*) that corresponds to a confidence level of 90%. This value can be obtained from a standard normal distribution table or calculated using statistical software.
Assuming a Z* value of 1.645 for a 90% confidence level, the margin of error is calculated as:
Margin of Error = Z* * √[(Sample Proportion * (1 - Sample Proportion)) / Sample Size]
Margin of Error = 1.645 * √[(0.4 * 0.6) / 500]
Margin of Error ≈ 0.0343
Now, we can calculate the confidence interval by adding and subtracting the margin of error from the sample proportion:
Confidence Interval = 0.4 ± 0.0343
Confidence Interval = (0.3657, 0.4343)
Therefore, the 90% confidence interval for the percentage of persons age 18 to 24 in the city who are attending college is approximately 36.57% to 43.43%.