Question 973393: According to a survey in a country, 38% of adults do not have any credit cards. Suppose a simple random sample of 900 adults is obtained.
A) Determine the mean of the sampling distribuiton of p.
B) Determine the standard deviation of the sampling distribution of p.
C) In a random sample of 900 adults, what is the probability that less than 36% have no credit cards?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The mean of the sampling distribution is the point estimate, which is 0.38 or 342 adults from 900.
The standard deviation is
sqrt {[(p)(1-p)]/(900)}
This is sqrt {(0.38)*(0.62)/900}
=sqrt (0.0002618)=0.0162. This is the standard deviation of the sampling distribution.
The z-value for the random sample is {0.36-0.38)/0.0162; it is the sample-the postulated mean all divided by the standard error of the sampling distribution.
This is (-.02/0.0162), and we want the area on the normal distribution to the left of z< -1.234
We would expect with a large sample to have a fairly small probability of being 2% away. As a rough guide, the error is 1/sqrt (n), here about 3%. But when the probability differs from 50%, which it does, the error is less.
Probability is 0.1085. A z-score of -1.28 is the 10th percentile, so this is reasonable.
|
|
|
