Question

For Americans using library services, the American Library Association claims that at most $67\%$ of patrons borrow books. The library director in Owensboro, Kentucky feels this is not true, so she asked a local college statistic class to conduct a survey. The class randomly selected $100$ patrons and found that $82$borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY? Use $\alpha =0.01$ level of significance. What is the possible proportion of patrons that do borrow books from the Owensboro Library?

Short answer

There is sufficient evidence to conclude that more than $67\%$of patrons borrow books in Owensboro, KY. The possible proportion of patrons that do borrow books from the library is $0.82$.

Step-by-step solution

Given information 

$x=82,n=100,p_o=0.67,a=0.01$

Explanation

State the hypothesis:

The null hypothesis states that most $67\%$of the patrons borrow books in Owensboro, KY. In symbols:

$H_o:p\le 0.67$

The alternative hypothesis states that more than $67\%$of patrons borrow books in Owensboro, KY. In symbols:

$H_a:p>0.67$

The sample proportion is:

$$\begin{gathered} \stackrel{⏜}{P}=\frac{x}{n} \\ =\frac{82}{100} \\ =0.82 \end{gathered}$$

The test statistic is:

$$\begin{gathered} z=\frac{\stackrel{⏜}{p}-p_0}{\sqrt{{\displaystyle \frac{p_0\left(1-p_0\right)}{n}}}} \\ =\frac{0.82-0.67}{\sqrt{{\displaystyle \frac{0.67\left(1-0.67\right)}{100}}}} \\ =\frac{0.150}{0.0470} \\ =3.19 \end{gathered}$$

For the right-tailed test, the p-value is:

$$\begin{gathered} p-value=P\left(z>3.19\right) \\ =0.0007 \end{gathered}$$

Since $p-value=0.0007<0.01$, reject the null hypothesis.

There is sufficient evidence to conclude that $51.7\%$of homes were heated by natural gas.