Question

1. The Central Limit Theorem for the proportion requires np >= 10 and n(1-p) >= 10, and that the sample was collected using an SRS. If these requirements are met, then the distribution of the sample proportion is approximately normal. If we know that the population proportion is p, what are the mean and standard deviation for the distribution of the sample proportion?

Assume you know that n = 1002 and p = .50. Use this information in problems 2-4.

2. What are the mean and standard deviation (round to four decimal places) for the sampling distribution of the sample proportion given the info above?

3. Based on the data above, construct and solve a z-score (round to two decimal places) for this problem using a sample proportion of .53.

4. Using the z-score value from above, what is the probability that the sample proportion is less than .53?

Answer 1

1) Given n = 1002 and p = 0.50

And also the requirements are met and sampling distribution of sample proportion is approximately normal.

So mean of sample proportion p^ = p = 0.50

Standard deviation of p^ = p*(1-p)/n = (0.50*0.50)/1002

  p^ = 0.0158

2) Mean p^ = 0.50

Standard deviation p^ = 0.0158

3) Z = p^ - p/p^

Z = 0.53 - 0.50/0.0158

Z = 1.90

4) P( p^ < 0.53 ) = P ( p^ - p /p^ < 0.53 - 0.50/0.0158)

= P( Z < 1.90)

Using Z table

P(p^ < 0.53 ) = 0.9713