Question

A random sample is to be selected from a population that has a proportion of successes p = 0.61. Determine the mean and standard deviation of the sampling distribution of p? for each of the following sample sizes. (Round your standard deviations to four decimal places.)
(a) n = 20

mean
standard deviation

(b) n = 30

mean
standard deviation

(c) n = 40

mean
standard deviation

(d) n = 60

mean
standard deviation

(e) n = 110

mean
standard deviation

(f) n = 210

mean
standard deviation

Answer 1

Answer with explanation:

We know that the mean and standard deviation of the sampling distribution of p is given by :-

$\mu_{\hat{p}}=p\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}$, here p is the proportion of successes ion and n is the sample size.

We are given that ,

Proportion of successes p = 0.61

a) n= 20

The mean and standard deviation of the sampling distribution of p is given by :-

$\mu_{\hat{p}}=0.61\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{0.61(1-0.61)}{20}}\\\\=\sqrt{0.011895}\approx0.1091$

So,  mean  = 0.61

standard deviation = 0.1091

(b) n = 30

$\mu_{\hat{p}}=0.61\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{0.61(1-0.61)}{30}}\\\\=\sqrt{0.00793}\approx0.0891$

mean

=0.61

standard deviation  =0.0891

(c) n = 40

$\mu_{\hat{p}}=0.61\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{0.61(1-0.61)}{40}}\\\\\approx0.0771$

mean

=0.61

standard deviation  =0.0771

(d) n = 60

 $\mu_{\hat{p}}=0.61\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{0.61(1-0.61)}{60}}\\\\\approx0.0630$

mean  =0.61

standard deviation  =0.0630

(e) n = 110

 $\mu_{\hat{p}}=0.61\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{0.61(1-0.61)}{110}}\\\\\approx0.0465$

mean  =0.61

standard deviation  = 0.0465

(f) n = 210

 $\mu_{\hat{p}}=0.61\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{0.61(1-0.61)}{210}}\\\\\approx0.0337$

mean  =0.61

standard deviation =0.0337