Question

Assume that a sample is used to estimate a population proportion p. Find the 95% confidence interval for a sample of size 393 with 37% successes. Enter your answer as an open interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

Answer 1

The 95% confidence interval for the population proportion with a sample size of 393 and 37% success is (0.3217, 0.4183).

Step-by-step explanation:

The student has provided a sample size (n) of 393 individuals with 37% reporting successes. To find the 95% confidence interval for the population proportion, we'll use the formula for the confidence interval of a population proportion, which is:

p±*z*((p*(1-p)/n)^0.5)

Where p is the sample proportion, z is the z-score corresponding to the 95% confidence level, and n is the sample size.

First, we calculate z by looking at the standard normal distribution (z-score for 95% confidence is approximately 1.96). Next, we convert the sample proportion of successes from percentage to decimal (37% = 0.37) and plug in the values:

p' = 0.37
q' = 1 - p' = 0.63
n = 393
z = 1.96

Now compute the margin of error (ME):

ME = z*((p*q/n)^0.5) = 1.96*((0.37*0.63/393)^0.5)

Using a calculator, we find that ME is approximately 0.0483.

Finally, create the confidence interval:

(p' - ME, p' + ME) = (0.37 - 0.0483, 0.37 + 0.0483) = (0.3217, 0.4183)

Therefore, the 95% confidence interval for the population proportion is (0.3217, 0.4183).