Chapter 7, Problem T7.13SPT

(a)

To determine

To Calculate: the probability that at least 29% of the sample are from poverty level households.

Answer to Problem T7.13SPT

0.17%

Explanation of Solution

Given:

  $\begin{array}{c}p=\text{22\%=0}\text{.22}\\ n\text{=5}0\\ \widehat{p}\text{=}0.29\end{array}$

Formula used:

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

  $z=\frac{x-\mu }{\sigma }$

Calculation:

  ${\mu }_{\widehat{p}}=p=22\%=0.22$

The standard deviation of the sampling distribution of the sample population $\widehat{p}$

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.22\left(1-0.22\right)}{300}}\\ =0.0239\end{array}$

The z-score is

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{0.29-0.22}{0.0239}\\ =2.93\end{array}$

The standard normal probability is

  $\begin{array}{c}P\left(\widehat{p}\ge 0.29\right)=P\left(Z>2.93\right)\\ =1-P\left(Z<2.93\right)\\ =1-0.9983\\ =0.0017\\ =0.17\%\end{array}$

(b)

To determine

To Explain: that there is convincing evidence that the percentage of children under the age of 6 living in households with incomes less than the official poverty level in this state is greater than the national value of 22%.

Answer to Problem T7.13SPT

Yes

Explanation of Solution

Given:

  $\begin{array}{c}p=\text{22\%=0}\text{.22}\\ n\text{=5}0\\ \widehat{p}\text{=}0.29\end{array}$

Formula used:

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

  $z=\frac{x-\mu }{\sigma }$

Calculation:

  ${\mu }_{\widehat{p}}=p=22\%=0.22$

The standard deviation of the sampling distribution of the sample population $\widehat{p}$ is

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.22\left(1-0.22\right)}{300}}\\ =0.0239\end{array}$

The z-score is

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{0.29-0.22}{0.0239}\\ =2.93\end{array}$

Standard normal probability

  $\begin{array}{c}P\left(\widehat{p}\ge 0.29\right)=P\left(Z>2.93\right)\\ =1-P\left(Z<2.93\right)\\ =1-0.9983\\ =0.0017\\ =0.17\%\end{array}$

A probability is observed to be small, when the probability is less than 0.05.

In this case, the probability is very small and therefore it is unlikely to get at least 29% housed holds. This then means that there is convincing evidence that the percentage of children under the age of six living in households with incomes less than the official poverty level in this state is above the national value of 22%.