Families in Poverty. In $2012$, $15.9\%$of all U.S. families had incomes below the poverty level, as reported by the U.S. Census Bureau in American Community Survey. During that same year, of $400$randomly selected Wyoming families, had incomes below the poverty level. At the $5\%$significance level, do the data provide sufficient evidence to conclude that, in$2012$, the percentage of families with incomes below the poverty level was lower among those living in Wyoming than among all U.S. families?

The significance level is $5\%$

The number of families is $n=400$

$p_0=0.159$

Income below the poverty level is$x=50$

The sample proportion is calculated as

$\hat{p}=\frac{x}{n}$

$\hat{p}=\frac{50}{400}$

$=0.125$

Calculate the value of $n{p}_0$

$n{p}_0=\left(400\right)(0.159)$

role="math" localid="1651415199813" $=63.6$

Calculate the value of

$n\left(1-p_0\right)=\left(400\right)(1-0.159)$

$=336.4$

Both the values are greater than $5$.

So one proportion $z-$test is appropriate to use.

The null hypothesis is

$H_0:p_0=0.159$

The alternate hypothesis is

$H_0:p_0<0.159$

The expression for $z$is

$z=\frac{\hat{p}-p_0}{\sqrt{\frac{P_0\left(1-p_0\right)}{n}}}$

The $z$value is calculated as

$z=\frac{0.125-0.159}{\sqrt{\frac{0.159(1-0.159)}{400}}}$

$=\frac{-0.034}{0.0183}$

$=-1.86$

When $\alpha =0.05$the tail is left tailed

The critical value of $z$from the standard table is

$z_{0.05}=-1.645$

The test statistic is in the negative range. As a result, the hypothesis $H_0$is ruled out.

At the$5\%$ level, the test results are statistically significant.

As a result, no, the data does not provide adequate evidence to conclude that the percentage of families with incomes below the poverty line was lower in Wyoming than in the rest of the United States.