According to government data, 22% of American children under the age of 6 live in households with incomes less than the official poverty level. A study of learning in early childhood chooses an SRS of 300 children from one state and finds that $\mathrm{p}\wedge \widehat{\mathrm{p}}=0.29$.

a. Find the probability that at least 29% of the sample are from poverty-level households, assuming that 22% of all children under the age of 6 in this state live in poverty-level households.

b. Based on your answer to part (a), is there 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%? Explain your reasoning.

The given probability is $\mathrm{p}=22\mathrm{\%}=0.22$

$\begin{array}{l}\mathrm{n}=50\\ \widehat{\mathrm{p}}=0.29\end{array}$

The following formula was used:

$\begin{array}{r}{\mathrm{\sigma }}_{\widehat{\mathrm{p}}}=\sqrt{\frac{\mathrm{p}(1-\mathrm{p})}{\mathrm{n}}}\\ \mathrm{z}=\frac{\mathrm{x}-\mathrm{\mu }}{\mathrm{\sigma }}\end{array}$

Consider that

${\mathrm{\mu }}_{\widehat{\mathrm{p}}}=\mathrm{p}=22\mathrm{\%}=0.22$

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

role="math" localid="1654353485299" $\begin{array}{c}{\mathrm{\sigma }}_{\widehat{\mathrm{p}}}=\sqrt{\frac{\mathrm{p}(1-\mathrm{p})}{\mathrm{n}}}\\ =\sqrt{\frac{0.22(1-0.22)}{300}}\\ =0.0239\end{array}$

The z-result is

role="math" localid="1654353515716" $\begin{array}{c}\mathrm{z}=\frac{\mathrm{x}-\mathrm{\mu }}{\mathrm{\sigma }}\\ =\frac{0.29-0.22}{0.0239}\\ =2.93\end{array}$

The normal probability formula is:

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

The given probability is $\mathrm{p}=22\mathrm{\%}=0.22$

$\begin{array}{l}\mathrm{n}=50\\ \widehat{\mathrm{p}}=0.29\end{array}$

The following formula was used:

$\begin{array}{r}{\mathrm{\sigma }}_{\widehat{\mathrm{p}}}=\sqrt{\frac{\mathrm{p}(1-\mathrm{p})}{\mathrm{n}}}\\ \mathrm{z}=\frac{\mathrm{x}-\mathrm{\mu }}{\mathrm{\sigma }}\end{array}$

Consider that

${\mathrm{\mu }}_{\widehat{\mathrm{p}}}=\mathrm{p}=22\mathrm{\%}=0.22$

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

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

The z-result is

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

The normal probability formula is:

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

When the likelihood is less than 0.05, the probability is said to be small.

In this situation, the chance is really low, therefore getting at least 29 percent housed holds is implausible. This implies there is strong evidence that the percentage of children under the age of six living in households with earnings below the official poverty level in this state is higher than the national average of 22%.