The proportion of a population with a characteristic of interest is p = 0.37. Find the mean and standard deviation of the sample proportion P^ obtained from random samples of size 125.
for large samples we have
so the mean is 0.37
yes
but what does q and n stand for ?
i think thats where im confused
\[ \text {For large samples, the sample proportion is approximately normally distributed, with mean}\\ \huge \mu_\hat{p}= p \\ \text{and standard deviation} \\ \huge \sigma_\hat{p}=\sqrt{\frac{pq}{n}}\]
q = 1 - p , and n is the size of your sample
q is the complement of p
so it would be sqrt(1-0.37)/125
\[\huge \sigma_\hat{p}=\sqrt{\frac{p(1-p)}{n}}\]
And the answer I got was 0.071?
Im not sure if that's right =0
\[\huge \sigma_\hat{p}=\sqrt{\frac{.37(.63)}{125}}\]
ohh you multiply the 0.37 and .63!
yes :)
so it would be .0432!
well nvm it would be 0.043! thank you so much for breaking down the formula to me!!
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