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Chapter 7: Problem 35

The president of Doerman Distributors, Inc., believes that \(30 \%\) of the firm's orders come from first-time customers. A random sample of 100 orders will be used to estimate the proportion of first-time customers. a. Assume that the president is correct and \(p=.30 .\) What is the sampling distribution of \(\bar{p}\) for this study? b. What is the probability that the sample proportion \(\bar{p}\) will be between .20 and \(.40 ?\) c. What is the probability that the sample proportion will be between .25 and \(.35 ?\)

Short Answer

Expert verified
a. \( \bar{p} \sim N(0.3, 0.0458) \); b. 0.9708; c. 0.7242.

Step by step solution

01

Determine the Sampling Distribution of Sample Proportion

Given that the probability (p) of an order being from a first-time customer is 0.3. The sample size, n, is 100. The sampling distribution of the sample proportion \( \bar{p} \) can be approximated by a normal distribution because the sample size is large. The mean of this distribution is \( p = 0.3 \) and the standard deviation is given by the formula:\[ \sigma_{\bar{p}} = \sqrt{ \frac{p(1-p)}{n} } = \sqrt{ \frac{0.3 \times 0.7}{100} } = 0.0458. \]Thus, \( \bar{p} \sim N(0.3, 0.0458) \).
02

Calculate Probability for Sample Proportion Between 0.20 and 0.40

To find the probability that the sample proportion \( \bar{p} \) falls between 0.20 and 0.40, we'll use the normal distribution approximation:1. Convert the boundaries to z-scores: - \( z_{0.20} = \frac{0.20 - 0.30}{0.0458} \approx -2.18 \) - \( z_{0.40} = \frac{0.40 - 0.30}{0.0458} \approx 2.18 \)2. Look up these z-scores in a standard normal distribution table: - The probability corresponding to \( z = -2.18 \) is approximately 0.0146. - The probability corresponding to \( z = 2.18 \) is approximately 0.9854.3. The probability \( P(0.20 < \bar{p} < 0.40) = 0.9854 - 0.0146 = 0.9708 \).
03

Calculate Probability for Sample Proportion Between 0.25 and 0.35

For this range, perform similar calculations with the boundaries 0.25 and 0.35:1. Convert the boundaries to z-scores: - \( z_{0.25} = \frac{0.25 - 0.30}{0.0458} \approx -1.09 \) - \( z_{0.35} = \frac{0.35 - 0.30}{0.0458} \approx 1.09 \)2. Look up these z-scores in a standard normal distribution table: - The probability corresponding to \( z = -1.09 \) is approximately 0.1379. - The probability corresponding to \( z = 1.09 \) is approximately 0.8621.3. The probability \( P(0.25 < \bar{p} < 0.35) = 0.8621 - 0.1379 = 0.7242 \).

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Key Concepts

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

Normal Distribution Approximation
The normal distribution is a fundamental concept in statistics. It is a bell-shaped distribution that is symmetric around the mean. When we deal with proportions in large samples, we can use a normal distribution approximation to make calculations simpler. This works well when certain conditions are met, like a large enough sample size.
  • The sample size should be large, generally greater than 30.
  • The conditions for the normal approximation to apply include: np 鈮� 5 and n(1-p) 鈮� 5, where n is the sample size and p is the population proportion.

In this exercise, the sample size is 100, and the proportion of interest is 0.3. Given these conditions, we can safely use the normal distribution approximation for \( \bar{p} \), the sample proportion. This allows us to easily compute the probability of observing a sample proportion within a certain range.
Sample Proportion
The sample proportion is a statistic that represents the proportion of interest for a particular sample. It is denoted as \( \bar{p} \). In this scenario, we're looking at the orders from first-time customers.
  • The sample proportion is computed as: \( \bar{p} = \frac{x}{n} \), where x is the number of successes (e.g., first-time orders) and n is the sample size.
  • It is crucial to understand that while \( \bar{p} \) gives insights into the sample at hand, what we're often interested in is making inferences about the population proportion \( p \).

In the exercise given, the president believes the population proportion \( p \) is 0.3, which is the basis for calculating the sampling distribution of \( \bar{p} \). This helps in determining how \( \bar{p} \) might behave if the study is repeated many times.
Z-Score Calculation
Z-scores are a way to convert a specific example, like a sample mean or proportion, into a standardized value. This helps to understand where a sample statistic lies relative to the expected average.
  • The Z-score formula for a sample proportion is: \( z = \frac{\bar{p} - p}{\sigma_{\bar{p}}} \).
  • \( \sigma_{\bar{p}} \), the standard deviation of the sample proportion, can be computed using the formula: \( \sigma_{\bar{p}} = \sqrt{\frac{p(1-p)}{n}} \).

In the context of the exercise, the task is to compute the probability that the sample proportion \( \bar{p} \) lies within specific bounds like 0.20 to 0.40. By calculating and using the Z-scores, we can determine these probabilities using the standard normal distribution table. This standardization helps in comparing values from different datasets on a similar scale.

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Most popular questions from this chapter

The Grocery Manufacturers of America reported that \(76 \%\) of consumers read the ingredients listed on a product's label. Assume the population proportion is \(p=.76\) and a sample of 400 consumers is selected from the population. a. Show the sampling distribution of the sample proportion \(\bar{p}\), where \(\bar{p}\) is the proportion of the sampled consumers who read the ingredients listed on a product's label. b. What is the probability that the sample proportion will be within 卤.03 of the population proportion? c. Answer part (b) for a sample of 750 consumers.

Lori Jeffrey is a successful sales representative for a major publisher of college textbooks. Historically, Lori obtains a book adoption on \(25 \%\) of her sales calls. Viewing her sales calls for one month as a sample of all possible sales calls, assume that a statistical analysis of the data yields a standard error of the proportion of .0625 a. How large was the sample used in this analysis? That is, how many sales calls did Lori make during the month? b. Let \(\bar{p}\) indicate the sample proportion of book adoptions obtained during the month. Show the sampling distribution of \(\bar{p}\) c. Using the sampling distribution of \(\bar{p},\) compute the probability that Lori will obtain book adoptions on \(30 \%\) or more of her sales calls during a one-month period.

A survey question for a sample of 150 individuals yielded 75 Yes responses, 55 No responses, and 20 No Opinions. a. What is the point estimate of the proportion in the population who respond Yes? b. What is the point estimate of the proportion in the population who respond No?

Suppose a simple random sample of size 50 is selected from a population with \(\sigma=10\) Find the value of the standard error of the mean in each of the following cases (use the finite population correction factor if appropriate) a. The population size is infinite. b. The population size is \(N=50,000\) c. The population size is \(N=5000\) d. The population size is \(N=500\).

A sample of 50 Fortune 500 companies (Fortune, April 14,2003 ) showed 5 were based in New York, 6 in California, 2 in Minnesota, and 1 in Wisconsin. a. Develop an estimate of the proportion of Fortune 500 companies based in New York. b. Develop an estimate of the number of Fortune 500 companies based in Minnesota. c. Develop an estimate of the proportion of Fortune 500 companies that are not based in these four states.
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