Question

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)

Answer 1

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

Step-by-step explanation:

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

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

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

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 :-

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

$(\hat{p})\sim N(0.10,\ \sqrt{0.0003})$

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

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