Assume that a sample is used to estimate a population proportion p. Find the 90% confidence interval for a sample of size 146 with 24% successes.

Question

Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places. 

C.I. = 

Answer 1

To find the 90% confidence interval for a sample of size 146 with 24% successes, we can use the following formula:

Confidence interval = sample proportion ± z* (standard error)

where z* is the critical value from the standard normal distribution at the 90% confidence level, and the standard error is calculated as:

standard error = sqrt[(sample proportion * (1 - sample proportion)) / sample size]

Substituting the given values, we get:

sample proportion = 0.24

sample size = 146

z* for a 90% confidence level = 1.645 (from standard normal distribution)

standard error = sqrt[(0.24 * (1 - 0.24)) / 146] = 0.048

Now we can calculate the confidence interval as:

Confidence interval = 0.24 ± 1.645 * 0.048

Confidence interval = (0.159, 0.321)

Therefore, we can say with 90% confidence that the true population proportion lies within the range of 0.159 to 0.321.

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