Question

Out of a random sample of $65$freshmen at State University, $31$students have declared a major. Use the “plus-four” method to find a $96$% confidence interval for the true proportion of freshmen at State University who have declared a major.

Short answer

A $96$% two-sided Cl estimate for the true proportion of freshmen at State University who have declared a major is $0.355$$\le$p $\le$$0.602$.

Step-by-step solution

Given Information

Out of a random sample of n=$65$freshmen at State University, x=$31$ students have declared a major.

Step 2: Explanation

If $\stackrel{\wedge }{p}$is the proportion of observations in a random sample of size n that belongs to a class of interest, an approximate $100$($1$-$\alpha$) % confidence interval on the proportion p of the population that belongs to this class is

$\hat{p}-z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\le p\le \hat{p}+z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ (1)

$\text{where}{z}_{\frac{\alpha }{2}}\text{is the upper}\frac{\alpha }{2}\text{percentage point of the standard normal distribution.}$

Out of a random sample of n=65 freshmen at State University, x=31 students have declared a major.

Explanation

If we use the plus-four method, than n'=n+$4$=$69$and x'=x+$2$=$33$. Therefore, the point estimate of the proportion of the population p is

$\hat{p}=\frac{x^{\mathrm{\prime }}}{n^{\mathrm{\prime }}}=\frac{33}{69}=0.478$ (2)

For $96$% two-sided confidence interval,

$\frac{\alpha }{2}=\frac{1-0.96}{2}=0.02$

and

$z_{\frac{\alpha }{2}}=2.05$

Therefore, from Equations ($1$) and ($2$) a $96$% two-sided CI estimate for the true proportion of freshmen at State University who have declared a major is

$0.478-2.05\sqrt{\frac{0.478(1-0.478)}{69}}\le p\le 0.478+2.05\sqrt{\frac{0.478(1-0.478)}{69}}$

which simplifies to

$0.355\le p\le 0.602$