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Chapter 7: Problem 24

Barron 's reported that the average number of weeks an individual is unemployed is 17.5 weeks (Barron \(s\), February 18,2008 ). Assume that for the population of all unemployed individuals the population mean length of unemployment is 17.5 weeks and that the population standard deviation is 4 weeks. Suppose you would like to select a random sample of 50 unemployed individuals for a follow-up study. a. Show the sampling distribution of \(\bar{x}\), the sample mean average for a sample of 50 unemployed individuals. b. What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within 1 week of the population mean? c. What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within \(1 / 2\) week of the population mean?

Short Answer

Expert verified
The sampling distribution has mean 17.5 and standard error 0.5657. Probabilities: b) 0.922, c) 0.597.

Step by step solution

01

Understanding the Sampling Distribution

The sampling distribution of the sample mean \(\bar{x}\) is the distribution of the average of a sample of size \(n\). It is centered around the population mean and has a standard deviation equal to the population standard deviation divided by the square root of the sample size \(n\).
02

Identify Key Parameters

Given: Population mean (\(\mu\)) = 17.5 weeks, population standard deviation (\(\sigma\)) = 4 weeks, and sample size (\(n\)) = 50.
03

Determine the Sampling Distribution Parameters

The sampling distribution of \(\bar{x}\) has a mean (\(\mu_{\bar{x}}\)) equal to the population mean, so \(\mu_{\bar{x}} = 17.5\). The standard deviation of the sampling distribution, also known as the standard error (\(\sigma_{\bar{x}}\)), is calculated as \(\sigma / \sqrt{n}\). Thus, \(\sigma_{\bar{x}} = 4 / \sqrt{50} = 0.5657\).
04

Calculate Probability for Part b

Part b asks for the probability that the sample mean is within 1 week of the population mean. This is equivalent to calculating \(P(16.5 < \bar{x} < 18.5)\). Perform a standard normal distribution transformation: convert \(16.5\) and \(18.5\) to their corresponding z-scores using \(z = (x - \mu_{\bar{x}}) / \sigma_{\bar{x}}\). For \(16.5\), \(z = (16.5 - 17.5) / 0.5657 = -1.767\); for \(18.5\), \(z = (18.5 - 17.5) / 0.5657 = 1.767\). Use a standard normal distribution table or calculator to find \(P(-1.767 < Z < 1.767) = 0.922\).
05

Calculate Probability for Part c

For part c, calculate the probability the sample mean is within 0.5 weeks of the population mean: \(P(17 < \bar{x} < 18)\). Find z-scores for 17 and 18: for 17, \(z = (17 - 17.5) / 0.5657 = -0.884\); for 18, \(z = (18 - 17.5) / 0.5657 = 0.884\). Use the standard normal distribution to find \(P(-0.884 < Z < 0.884) = 0.597\).

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Key Concepts

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

Understanding Population Mean
The population mean is a fundamental concept in statistics that represents the average value of a particular characteristic for every member of a specified population. In this exercise, we see the population mean for unemployment being 17.5 weeks. This means when all unemployed individuals are considered, they spend an average of 17.5 weeks without employment.

When working with statistical data, the population mean provides a baseline that researchers and statisticians can compare to sample data gathered from the population. By comparing sample means to the population mean, we can make inferences about the population without surveying every member, saving time and resources.

The key takeaway is that the population mean acts as a summary measure, providing insights about the central tendency of the entire group being studied.
Importance of Standard Deviation
The standard deviation is a statistic that measures the dispersion or spread of a set of values. For the population of unemployed individuals, the standard deviation is given as 4 weeks. This value indicates how much the duration of unemployment varies from the population mean.

Knowing the standard deviation helps us understand how spread out the individuals' unemployment durations are from the average. A smaller standard deviation would imply that the weeks of unemployment are relatively close to the mean, while a larger one indicates more variability.
  • Standard deviation provides insight into the variability of data.
  • It helps in calculating the standard error when working with sample data.
For this exercise, the standard deviation plays a crucial role in finding the standard error of the sample mean, which tells us the expected difference between the sample mean and the population mean. The standard error is calculated by dividing the population standard deviation by the square root of the sample size, helping in probability assessments for samples.
Calculating Probability
Probability calculations in this context involve determining the likelihood that a sample mean falls within a specified range of the population mean. This exercise highlights using z-scores and the standard normal distribution to find probabilities related to the sample means.

Here's a brief breakdown of the steps involved:
  • Identify the range within which you want to calculate the probability. For example, being within 1 week or 0.5 weeks of the population mean.
  • Use the formula for calculating z-scores: \(z = \frac{x - \mu_{\bar{x}}}{\sigma_{\bar{x}}}\), where \(x\) is the value you are investigating, \(\mu_{\bar{x}}\) is the mean of the sampling distribution, and \(\sigma_{\bar{x}}\) is the standard error.
  • Look up the calculated z-score in standard normal distribution tables or use a calculator to find the probability.
  • The result tells you how likely it is for the sample mean to lie within the set range of the population mean.
This method is crucial for understanding how sample data may compare to population parameters, providing insights that are foundational to statistical inference.

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Most popular questions from this chapter

BusinessWeek conducted a survey of graduates from 30 top MBA programs ( Business Week , September 22,2003 ). On the basis of the survey, assume that the mean annual salary for male and female graduates 10 years after graduation is \(\$ 168,000\) and \(\$ 117,000,\) respectively. Assume the standard deviation for the male graduates is \(\$ 40,000,\) and for the female graduates it is \(\$ 25,000\) a. What is the probability that a simple random sample of 40 male graduates will provide a sample mean within \(\$ 10,000\) of the population mean, \(\$ 168,000 ?\) b. What is the probability that a simple random sample of 40 female graduates will provide a sample mean within \(\$ 10,000\) of the population mean, \(\$ 117,000 ?\) c. In which of the preceding two cases, part (a) or part (b), do we have a higher probability of obtaining a sample estimate within \(\$ 10,000\) of the population mean? Why? d. What is the probability that a simple random sample of 100 male graduates will provide a sample mean more than \(\$ 4000\) below the population mean?

The mean annual cost of automobile insurance is \(\$ 939\) (CNBC, February 23,2006 ). Assume that the standard deviation is \(\sigma=\$ 245\) a. What is the probability that a simple random sample of automobile insurance policies will have a sample mean within \(\$ 25\) of the population mean for each of the following sample sizes: \(30,50,100,\) and \(400 ?\) b. What is the advantage of a larger sample size when attempting to estimate the population mean?

Americans have become increasingly concerned about the rising cost of Medicare. In 1990 the average annual Medicare spending per enrollee was \(\$ 3267 ;\) in \(2003,\) the average annual Medicare spending per enrollee was \(\$ 6883\) (Money, Fall 2003 ). Suppose you hired a consulting firm to take a sample of fifty 2003 Medicare enrollees to further investigate the nature of expenditures. Assume the population standard deviation for 2003 was \(\$ 2000\). a. Show the sampling distribution of the mean amount of Medicare spending for a sample of fifty 2003 enrollees. b. What is the probability that the sample mean will be within \(\pm \$ 300\) of the population mean? c. What is the probability that the sample mean will be greater than \(\$ 7500 ?\) If the consulting firm tells you the sample mean for the Medicare enrollees it interviewed was \(\$ 7500,\) would you question whether the firm followed correct simple random sampling procedures? Why or why not?

Assume that the population proportion is \(.55 .\) Compute the standard error of the proportion, \(\sigma_{\bar{p}},\) for sample sizes of \(100,200,500,\) and \(1000 .\) What can you say about the size of the standard error of the proportion as the sample size is increased?

Roper ASW conducted a survey to learn about American adults' attitudes toward money and happiness (Money, October 2003). Fifty-six percent of the respondents said they balance their checkbook at least once a month. a. Suppose a sample of 400 American adults were taken. Show the sampling distribution of the proportion of adults who balance their checkbook at least once a month. b. What is the probability that the sample proportion will be within 卤.02 of the population proportion? c. What is the probability that the sample proportion will be within 卤.04 of the population proportion?
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