Chapter 7, Problem T7.7SPT

To determine

To find: the sampling distribution of the sample proportion would be

Answer to Problem T7.7SPT

Approximately Normal with mean 0.55 and standard deviation 0.03

Explanation of Solution

Given:

  $\begin{array}{c}p=55\%=0.55\\ n=250\end{array}$

Formula used:

  ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

Calculation:

The sample proportion is approximately normal, if $np\ge 10$ and $n\left(1-p\right)\ge 10$

  $\begin{array}{c}n{p}_0=250\left(0.55\right)=137.5\ge 10\\ n\left(1-p_0\right)=250\left(1-0.55\right)=112.5\ge 10\end{array}$

  ${\mu }_{\widehat{p}}=p=0.55$

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.55\left(1-0.55\right)}{250}}\\ =0.03\end{array}$

Hence, the correct option is (b)