Question

Suppose that a population parameter is 0.9 and many large samples are taken from the population. If the sample proportions are normally distributed, with 99.7% of the sample proportions falling between 0.867 and 0.933, what is the standard deviation of the sample proportions?

A. 0.025
B. 0.022
C. 0.011
D. 0.015

Answer 1

Well the formula for standard deviation is difficult but it should be B

Answer 2

So if I plugged those numbers into the standard deviation formula I would get 0.022 as my answer?

Answer 3

Yes, most likely

Answer 4

@perl could you explain this to me if you don't mind I am confused as to how you solve this problem.

Answer 5

the standard deviation of the sampling distribution of the sample proportion is
$$ \Large \rm { \sigma_p= \sqrt{\frac{p (1-p)}{n}}, \\~\\
~This~ is~valid~ when ~population ~is~ ten~ times\\~ bigger ~ than ~ sample.



}

$$

Answer 6

* standard deviation of the sampling distribution of sample proportions of size n

Answer 7

what numbers do I plug into where in the formula

Answer 8

we want the critical score for 99.7% significance level
but 99.7% = .997
we need to find the z score for this and i found it to be 2.9677