91Ó°ÊÓ

Given:

$$\begin{gathered} p=22\%=0.22 \\ n=50 \\ \hat{p}=0.29 \end{gathered}$$

Formula used:

$$\begin{gathered} {\mathrm{σ}}_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \\ z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \end{gathered}$$

Concider,${\mathrm{μ}}_{\hat{p}}=p=22\%=0.22$

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

$$\begin{gathered} {\mathrm{σ}}_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \\ =\sqrt{\frac{0.22(1-0.22)}{300}} \\ =0.0239 \end{gathered}$$

The z-score is

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{0.29-0.22}{0.0239} \\ =2.93 \end{gathered}$$

The standard normal probability is

$$\begin{gathered} P(\hat{p}â‰\yen 0.29)=P(Z>2.93) \\ =1-P(Z<2.93) \\ =1-0.9983 \\ =0.0017 \\ =0.17\% \end{gathered}$$

Given:

Formula used:

$$\begin{gathered} {\mathrm{σ}}_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \\ z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \end{gathered}$$

Consider,${\mathrm{μ}}_{\hat{p}}=p=22\%=0.22$

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

$$\begin{gathered} {\mathrm{σ}}_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \\ =\sqrt{\frac{0.22(1-0.22)}{300}} \\ =0.0239 \end{gathered}$$

The z-score is

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{0.29-0.22}{0.0239} \\ =2.93 \end{gathered}$$

Standard normal probability

$$\begin{gathered} P(\hat{p}â‰\yen 0.29)=P(Z>2.93) \\ =1-P(Z<2.93) \\ =1-0.9983 \\ =0.0017 \\ =0.17\% \end{gathered}$$

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 $22\%$.