The US Department of Energy reported that $51.7\%$ of homes were heated by natural gas. A random sample of $221$homes in Kentucky found that $115$ were heated by natural gas. Does the evidence support the claim for Kentucky at the $\alpha =0.05$ level in Kentucky? Are the results applicable across the country? Why?

$x=115,n=221,p_0=0.517,a=0.05$

State the hypothesis:

The null hypothesis states that $51.7\%$of homes were heated by natural gas. In symbols:

$H_0:p=0.517$

The alternative hypothesis states that not $51.7\%$of homes were heated by natural gas. In symbols:

$H_a:p\ne 0.517$

The sample proportion is:

$$\begin{gathered} \stackrel{⏜}{P}=\frac{x}{n} \\ =\frac{115}{221} \\ =0.520 \end{gathered}$$

The test statistic is:

$$\begin{gathered} z=\frac{\stackrel{⏜}{p}-p_0}{\sqrt{{\displaystyle \frac{p_0\left(1-p_0\right)}{n}}}} \\ =\frac{0.520-0.517}{\sqrt{{\displaystyle \frac{0.517\left(1-0.517\right)}{221}}}} \\ =\frac{0.003}{0.0336} \\ =0.10 \end{gathered}$$

The p-value is:

$$\begin{gathered} p-value=P\left(z<-0.10\right)+P\left(z>0.10\right) \\ =0.4602+0.4602 \\ =0.9204 \end{gathered}$$

Since $p-value=0.9203>0.05$, do not reject the null hypothesis. There is sufficient evidence to conclude that $51.7\%$of homes were heated by natural gas. The results are not applicable across the country because the sample was selected from Kentucky only and may not be representative of the country.