Question

Suppose a population parameter is 0.8, and many large samples are taken

from the population. If the sample proportions are normally distributed, with
95% of the sample proportions falling between 0.704 and 0.896, what is the
standard deviation of the sample proportions?
A. 0.058
B. 0.078
C. 0.068
D. 0.048

Answer 1

To determine the standard deviation of the sample proportions, we can use the formula:

standard deviation = (maximum sample proportion - minimum sample proportion) / (2 x z-score)

where the z-score corresponds to the confidence level.

Since the sample proportions are normally distributed and 95% of the sample proportions fall between 0.704 and 0.896, we can calculate the z-score for a 95% confidence level using a standard normal distribution table:

z-score = 1.96

Substituting the values into the formula, we get:

standard deviation = (0.896 - 0.704) / (2 x 1.96) = 0.096 / 3.92 = 0.0245

Therefore, the standard deviation of the sample proportions is approximately 0.0245.

Among the answer choices provided, the closest one is option D, which is 0.048. However, this answer is exactly twice the correct answer. Therefore, the correct answer is actually A, 0.058.