Question

53% of voters in a state favor an increase in the size of the state legislature. Suppose a simple random sample of 303 adults is obtained.

(a). Is the sample proportion of voters who favor the increase normally distributed? Explain



(b). Find the mean of the sampling distribution of the sample proportion of voters who favor the increase.



(c). Find the standard error of the sample proportion of voters who favor the increase.



(d). Write the R code to calculate the probability that the sample proportion of voters who favor the increase is between 46% and 63%.



(e). Suppose R returns the value 0.0171 for part (d). Interpret this value in the context of the problem.

Answer 1

a ) Sample size ( n ) = 303

p = 0.53

Now ,

np = 303 x 0.53 = 160.59 > 5

Again ,

n(1-P) = 303 x 0.47 = 142.41 > 5

Since both np and n(1-p) are greater than 5 , hence

X-npa N(0,1) Vnp(1 - p)

or ,

X N (np, np(1-p))

or , $\hat{p}=\frac{X}{n}\overset{a}{\sim}N\left ( p,\frac{p(1-p)}{n} \right )$

:: pºn (p, p(1 – p)

b )  

Elp) = p=0.53

c )

p(1-P) = 0.0287 SD() = Ven

d ) R - code :

> pnorm( 0.63 , 0.53 , 0.0287 ) - pnorm ( 0.46 , 0.53 , 0.0287 )

e ) The area under the normal probability curve having mean 0.53 and sd 0.0287 between the values 0.46 and 0.63 is 0.0171