Question

Suppose that a population parameter is 0.3 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 0.018 and 0.582, what is the standard deviation of the sample proportions?

A. 0.094
B. 0.074
C. 0.064
D. 0.084

Answer 1

@perl I am following the steps that you gave me last week for a question like this but how do I find the z score.

Answer 2

http://openstudy.com/study?source=twitter#/updates/552f320ae4b0863033976526

Answer 3

$$ \Large \rm {
Margin ~of ~ Error = Z_c \cdot \sqrt{\frac{p(1-p)}{n}}


}$$

Answer 4

the margin of error is half of the given interval

Answer 5

is p the 0.3?

Answer 6

yes

Answer 7

what is n then

Answer 8

.997?

Answer 9

n is not given

Answer 10

n is the sample size

Answer 11

but you know margin of error, and you know Zc

Answer 12

so half of 0.3 is 0.15

Answer 13

$$
\Large \rm {

Margin ~of ~ Error = Z_c \cdot \sqrt{\frac{p(1-p)}{n}}
\\~\\Margin ~of ~ Error = \frac{0.582 - 0.018}{2}
\\ Z_{99.7} = 3

}

$$

Answer 14

0.282

Answer 15

so it turns out you dont need to know n, since the right hand side is the standard deviation

Answer 16

A. 0.094 so all you really have to do is subtract the sample proportions divide by 2 to get 0.282 and divide over 3. Where do you get the 0.7?

Answer 17

$$\Large \rm {


Margin ~of ~ Error = Z_c \cdot \color{red}{Standard~Deviation}
\\Margin ~of ~ Error = Z_c \cdot \color{red}{\sqrt{\frac{p(1-p)}{n}}

}
\\~\\\large Margin ~of ~ Error = \frac{0.582 - 0.018}{2}= 0.282
\\ Z_{99.7} = 3
\\~\\0.282 = 3 \cdot \color{red}{\sqrt{\frac{0.3\cdot 0.7}{n}}}
\\\frac{0.282}{3} = \color{red}{\sqrt{\frac{0.3\cdot 0.7}{n}}}
}
$$

Answer 18

I put the standard deviation in red

Answer 19

if p = .3
then 1- p = .7

Answer 20

Ok I see

Answer 21

I got 0.094 as the answer thank you for explaining this to me again I still was not 100% sure how to solve these types of problems. Thank you :)