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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
In this problem, we determined there is approximately a 57.2% chance that the estimate would be within 0.05 of the actual population proportion of students at a particular college who would rent textbooks. We used the sampling distribution of the sample proportion (瘸) and the Normal distribution to calculate this probability.

Step by step solution

01

Identify the parameters and sample size

First, let's identify the given information: The population proportion (p) = 0.38 The sample size (n) = 100
02

Determine the mean and standard deviation

For the sampling distribution of the sample proportion, the mean (碌) is given by: 碌 = p So in this case, 碌 = 0.38. The standard deviation (蟽) of the sampling distribution is given by: 蟽 = \(\sqrt{\frac{p(1-p)}{n}}\) Plugging in the values: 蟽 = \(\sqrt{\frac{0.38(1-0.38)}{100}}\)
03

Calculate the standard deviation

After we plug in the values for p and n to the formula for the standard deviation. 蟽 = \(\sqrt{\frac{0.38(1-0.38)}{100}}\) = \(\sqrt{\frac{0.38 \times 0.62}{100}}\) = \(\sqrt{0.003956}\) 蟽 鈮� 0.0629
04

Check if the normal distribution is applicable

Before we continue, we need to check if the normal distribution is applicable to this problem. This can be checked with the following conditions: 1. np 鈮�10 2. n(1-p) 鈮�10 Checking the conditions: 1. np = 100 脳 0.38 = 38 鈮�10 2. n(1-p) = 100 脳 0.62 = 62 鈮� 10 Both conditions are met so we can apply the normal distribution.
05

Calculate the probabilities using the normal distribution

Now, we are ready to determine the probability that the estimate would be within 0.05 of the actual population proportion. For that, we can set up the following inequalities: 0.38 - 0.05 鈮� 瘸 鈮� 0.38 + 0.05 0.33 鈮� 瘸 鈮� 0.43 To find the probabilities, we'll use the Z-Score formula: Z = \(\frac{\hat{p} - 碌}{蟽}\) Lower limit probability (0.33): Z = \(\frac{0.33 - 0.38}{0.0629}\) 鈮� -0.7937 Upper limit probability (0.43): Z = \(\frac{0.43 - 0.38}{0.0629}\) 鈮� 0.7937 Using a Z-Score table, we find the probabilities: P(Z < -0.7937) 鈮� 0.2140 P(Z < 0.7937) 鈮� 0.7860
06

Calculate the probability of the estimate being within 0.05 of the actual proportion

Now we subtract the lower probability from the upper probability to find the final result: P(0.33 鈮� 瘸 鈮� 0.43) = P(Z < 0.7937) - P(Z < -0.7937) = 0.7860 - 0.2140 = 0.5720 So, there is approximately a 57.2% chance that the estimate would be within 0.05 of the actual population proportion.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Population Proportion
Understanding the concept of population proportion is crucial for statistical analysis, particularly when examining characteristics within a group. The population proportion, denoted as \(p\), represents the fraction or percentage of a population that exhibits a certain trait or characteristic. In the context of our exercise, \(38\%\) or \(0.38\) of the college students would rent textbooks if they could. This figure is the population proportion that the researchers aim to estimate using a sample.

When researchers use the sampling method, they select a smaller group from the population which should reflect the characteristics of the entire group accurately. The sample proportion, notated as \(\hat{p}\), is the estimated proportion based on this sample. By assessing \(\hat{p}\), the researchers predict the behavior or attributes of the whole population. However, sampling comes with uncertainty and this is where the concept of standard deviation and the normal distribution come into play to quantify this variability.
Standard Deviation
The standard deviation is a critical statistic that measures the amount of variation or dispersion from the average. In simpler terms, it indicates how spread out the values in a data set are. A smaller standard deviation points to values being close to the mean, while a larger standard deviation signals a wide range of values.

In the example at hand, we calculate the standard deviation for the sampling distribution of \(\hat{p}\), which is expressed with the formula \(\sigma = \sqrt{\frac{p(1-p)}{n}}\). Plugging in our population proportion (\(0.38\)) and sample size (\(100\)), we determine that \(\sigma \approx 0.0629\). This provides us with crucial information about the expected variation in \(\hat{p}\) from sample to sample. With a smaller standard deviation, we would anticipate sample proportions to be closer to the population proportion.
Normal Distribution
The normal distribution, often described as a bell curve due to its shape, is a probability distribution that is symmetric around the mean. It shows that data near the mean are more frequent in occurrence than data far from the mean. In practice, the normal distribution is used to model many random variables because of its convenience and mathematical properties.

In our textbook example, we assume that the sampling distribution of \(\hat{p}\) is approximately normally distributed, given that the sample size is large enough and the conditions \(np \geq 10\) and \(n(1-p) \geq 10\) are satisfied. These conditions ensure that there are a sufficient number of 'successes' and 'failures' in the sample, which allow the sample distribution to take on that bell curve shape. Using the properties of the normal distribution, we can then calculate the probability that the sample proportion falls within a certain range of the population proportion, allowing researchers to assess the reliability of their estimates.

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Most popular questions from this chapter

Suppose that a particular candidate for public office is favored by \(48 \%\) of all registered voters in the district. A polling organization will take a random sample of 500 of these voters and will use \(\hat{p},\) the sample proportion, to estimate \(p\). a. Show that \(\sigma_{\hat{p}}\), the standard deviation of \(\hat{p},\) is equal to 0.022 . b. If for a different sample size, \(\sigma_{j}=0.071,\) would you expect more or less sample-to-sample variability in the sample proportions than when \(n=500 ?\) c. Is the sample size that resulted in \(\sigma_{\hat{p}}=0.071\) larger than 500 or smaller than \(500 ?\) Explain your reasoning.

Consider the following statement: The Department of Motor Vehicles reports that the proportion of all vehicles registered in California that are imports is \(0.22 .\) a. Is the number that appears in boldface in this statement a sample proportion or a population proportion? b. Which of the following use of notation is correct, \(p=0.22\) or \(\hat{p}=0.22 ?\) (Hint: See definitions and notation on page \(403 .\) )

For which of the following sample sizes would the sampling distribution of \(\hat{p}\) be approximately normal when $$ \begin{array}{c} p=0.2 ? \text { When } p=0.8 ? \text { When } p=0.6 ? \\ n=10 \quad n=25 \\ n=50 \quad n=100 \end{array} $$

Consider the two relative frequency histograms at the top of the next page. The histogram on the left was constructed by selecting 100 different random samples of size 40 from a population consisting of \(20 \%\) part-time students and \(80 \%\) full-time students. For each sample, the sample proportion of part-time students, \(\hat{p},\) was calculated. The \(100 \hat{p}\) values were used to construct the histogram. The histogram on the right was constructed in a similar way, but using samples of size 70 . a. Which of the two histograms indicates that the value of \(\hat{p}\) has smaller sample-to-sample variability? How can you tell? b. For which of the two sample sizes, \(n=40\) or \(n=70,\) do you think the value of \(\hat{p}\) would be less likely to be close to \(0.20 ?\) What about the given histograms supports your choice?

Explain what it means when we say the value of a sample statistic varies from sample to sample.
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