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The proportion of left handed people in the population is about 0.10. Suppose a random sample of 300 people is observed. (a) What is the sampling distribution of the sample proportion (p-hat)

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Answer:
(\hat{p})\sim N(0.10,\ 0.017)

Explanation:

Sampling distribution of the sample proportion
(\hat{p}) :


(\hat{p})\sim N(p,\ \sqrt{(p(1-p))/(n)})

The sampling distribution of the sample proportion
(\hat{p}) has mean =
\mu_{\hat{p}}=p and standard deviation =
\sigma_{\hat{p}}=\sqrt{(p(1-p))/(n)}.

Given : The proportion of left handed people in the population is about 0.10.

i.e. p=0.10

sample size : n= 300

Then , the sampling distribution of the sample proportion
(\hat{p}) will be :-


(\hat{p})\sim N(0.10,\ \sqrt{(0.10(1-0.10))/(300)})


(\hat{p})\sim N(0.10,\ √(0.0003))


(\hat{p})\sim N(0.10,\ 0.017) (approx)

Hence, the sampling distribution of the sample proportion
(\hat{p}) is
(\hat{p})\sim N(0.10,\ 0.017)

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