Question

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 ${\mathrm{p}}^{\wedge }\hat{p}=$.

a. Find the probability that at least $29\%$of the sample are from poverty-level households$0.29$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.

Short answer

(a)The standard normal probability is $0.17\%$

(b) the official poverty level in this state is above the national value of $22\%$

Step-by-step solution

Part (a) Step 1: Given Information

Given:

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

Formula used:

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

Part (a) Step 2: Simplification

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

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

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

The z-score is

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

The standard normal probability is

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

Part (b) Step 1: Given Information

Given:

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

Formula used:

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

Part (b) Step 2: Simplification

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

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

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

The z-score is

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

Standard normal probability

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

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 $22\%$.