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 α = 0.01 level of significance. What is the possible proportion of patrons that do borrow books from the Owensboro Library?

Short answer

CI: (0.7447, 0.8953)

Step-by-step solution

state the null and alternate hypothesis. we have to conduct hypothesis test if American using library services take more than 67% of patrons to borrow books according to American Library Association, on average. 

$H_0:p\le 0.67H_a:p>0.67$

Calculate the p-value using the normal distribution for proportions: 

p-value = 0.0007

In one to two complete sentences, explain what the p-value means for this problem. If the null hypothesis is true (the proportion is 0.67), then there is a 0.0007 probability that the sample (estimated) proportion is 0.82 or more.

Compare α and the p-value:Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences.

alpha	decision	reason for decision
0.01	Reject the null hypothesis.	$p-value<0.01$

Conclusion: At the 1% significance level, there is enough evidence to conclude that the proportion of patrons in Owensboro, KY who borrow books is more than 0.67. The class demonstrated that the percentage was higher than 67%.

Confidence interval and true proportion

To determine the possible proportion of patrons who do borrow books, construct a 95% confidence interval. It is (0.7447, 0.8953). We are 95% confident that the true population proportion of patrons in Owensboro, KY who borrow books is between 0.7447 and 0.8953 (between 74.47% and 89.53%).