Question

Some colleges now allow students to rent textbooks for a semester. Suppose that \(38 \%\) of all students enrolled at a particular college would rent textbooks if that option were available to them. If the campus bookstore uses a random sample of size 100 to estimate the proportion of students at the college who would rent textbooks, is it likely that this estimate would be within 0.05 of the actual population proportion? Use what you know about the sampling distribution of \(\hat{p}\) to support your answer.

Short answer

Yes/No, it is/isn't likely that the sample proportion estimate would be within 0.05 of the actual population proportion, depending on whether the calculated margin of error was less than, equal to or greater than 0.05.

Step-by-step solution

Calculation of Standard Deviation for the proportion

The standard deviation for the proportion \(\hat{p}\) can be computed using the formula: \[ \sqrt{\frac{{p(1-p)}}{n}} \] where \(p = 0.38\) is the population proportion and \(n = 100\) is the sample size. So, the standard deviation (SD) will be: \[ SD = \sqrt{\frac{{0.38(1-0.38)}}{100}}\]

Calculation of margin of error

We know that a normal rule of thumb is to use a z-score of 1.96 for a 95% confidence interval. The margin of error (ME) is then calculated using the formula: \[ ME = Z * SD \] where \(Z = 1.96\) and SD is the standard deviation calculated in the previous step.

Comparison of margin of error with required precision

The requested precision in the problem statement is 0.05. If the calculated margin of error (ME) from the previous step is less than or equal to 0.05, then it is likely that the sample proportion estimate would be within 0.05 of the actual population proportion.