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Chapter 8: Problem 37

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

Expert verified
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

01

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: \[ SD = \sqrt{\frac{{0.38(1-0.38)}}{100}}\]
02

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.
03

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.

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Most popular questions from this chapter

Explain what the term sampling variability means in the context of using a sample proportion to estimate a population proportion.

A random sample of 50 registered voters in a particular city included 32 who favored using city funds for the construction of a new recreational facility. For this sample, \(\hat{p}=\frac{32}{50}=\) 0.64 . If a second random sample of 50 registered voters was selected, would it surprise you if \(\hat{p}\) for that sample was not equal to 0.64 ? Why or why not?

Explain why there is sample-to-sample variability in \(\hat{p}\) but not in \(p\).

"Tongue Piercing May Speed Tooth Loss, Researchers Say" is the headline of an article that appeared in the San Luis Obispo Tribune (June 5, 2002). The article describes a study of 52 young adults with pierced tongues. The researchers believed that it was reasonable to regard this sample as a random sample from the population of all young adults with pierced tongues. The researchers found receding gums, which can lead to tooth loss, in 18 of the study participants. a. Suppose you are interested in learning about the value of \(p,\) the proportion of all young adults with pierced tongues who have receding gums. This proportion can be estimated using the sample proportion, \(\hat{p} .\) What is the value of \(\hat{p}\) for this sample? b. Based on what you know about the sampling distribution of \(\hat{p}\), is it reasonable to think that this estimate is within 0.05 of the actual value of the population proportion? Explain why or why not. (Hint: See Example 8.4\()\)

A random sample is to be selected from a population that has a proportion of successes \(p=0.25\). Determine the mean and standard deviation of the sampling distribution of \(\hat{p}\) for each of the following sample sizes: a. \(n=10\) d. \(n=50\) b. \(n=20\) e. \(n=100\) c. \(n=30\) f. \(n=200\)
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