/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Income of farm workers For the p... [FREE SOLUTION] | 91Ó°ÊÓ

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Income of farm workers For the population of farm workers in New Zealand, suppose that weekly income has a distribution that is skewed to the right with a mean of \(\mu=\$ 500(\mathrm{~N} . Z .)\) and a standard deviation of \(\sigma=\$ 160 .\) A researcher, unaware of these values, plans to randomly sample 100 farm workers and use the sample mean annual income \(\bar{x}\) to estimate \(\mu\). a. Show that the standard deviation of \(\bar{x}\) equals 16.0 . b. Explain why it is almost certain that the sample mean will fall within \(\$ 48\) of \(\$ 500\). c. The sampling distribution of \(\bar{x}\) provides the probability that \(\bar{x}\) falls within a certain distance of \(\mu,\) regardless of the value of \(\mu\). Show how to calculate the probability that \(\bar{x}\) falls within \(\$ 20\) of \(\mu\) for all such workers. (Hint: Using the standard deviation, convert the distance 20 to a \(z\) -score for the sampling distribution.)

Short Answer

Expert verified
a. \( \sigma_{\bar{x}} = 16 \). b. \( \bar{x} \) is almost certain to be within \( \$48 \) of \( \mu \) because of the Empirical Rule. c. Probability that \( \bar{x} \) is within \( \$20 \) of \( \mu \) is about 78.88%.

Step by step solution

01

Understanding the Standard Deviation of the Sample Mean

To find the standard deviation of the sample mean \( \bar{x} \), we use the formula: \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \). Here, \( \sigma = 160 \) and \( n = 100 \). Plug these values into the formula to get the standard deviation of the sample mean.
02

Calculating the Standard Deviation

Substitute \( \sigma = 160 \) and \( n = 100 \) into the formula for the standard deviation of the sample mean: \( \sigma_{\bar{x}} = \frac{160}{\sqrt{100}} = \frac{160}{10} = 16 \). Thus, the standard deviation of \( \bar{x} \) is 16.
03

Understanding the Range of the Sample Mean

The Central Limit Theorem tells us that the sampling distribution of \( \bar{x} \) will be approximately normal for large samples. Since the standard deviation of \( \bar{x} \) is 16, most sample means will fall within \( 3 \times 16 = 48 \) units of the population mean \( 500 \) due to the properties of a normal distribution, specifically the Empirical Rule.
04

Using the Empirical Rule

The Empirical Rule states that for a normal distribution, about 99.7% of the data falls within 3 standard deviations from the mean. This means that \( \bar{x} \) is almost certainly within \( 500 \pm 48 \), since 48 is 3 times the standard deviation of 16.
05

Converting Distance to a Z-score

To determine the probability that \( \bar{x} \) falls within \( \$20 \) of \( \mu \), convert \( 20 \) to a \( z \)-score using \( z = \frac{\text{distance}}{\sigma_{\bar{x}}} \). Substitute \( \text{distance} = 20 \) and \( \sigma_{\bar{x}} = 16 \): \( z = \frac{20}{16} = 1.25 \).
06

Calculating the Probability

Using standard normal distribution tables or a calculator, find the probability that corresponds to \( z = 1.25 \). This gives the probability that the sample mean falls within 20 dollars of the population mean. The cumulative probability up to \( z=1.25 \) is about 0.8944. Thus, the probability that \( \bar{x} \) is within 20 dollars of \( \mu \) is \( 0.8944 - (1 - 0.8944) = 0.7888 \), or about 78.88%.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental concept in statistics. It explains how, for sufficiently large sample sizes, the distribution of the sample mean will approximate a normal distribution. This happens even if the original population distribution is not normal. In our problem with farm worker incomes that are skewed to the right, the CLT assures us that the sample mean distribution will still be approximately normal as long as the sample size is large enough, like the sample of 100 workers here.
\[ \bar{x} \sim N(\mu, \frac{\sigma^2}{n}) \]

Because of this theorem, we can confidently predict the behavior of the sample mean, which means we can use tools and properties of the normal distribution, such as the Empirical Rule, and convert distances to z-scores for calculations, despite the original skew in the data.
The CLT is essential in making inferences about a population mean based on a sample mean, as it allows us to estimate probabilities and make predictions about where the sample mean is likely to fall relative to the population mean.
Empirical Rule
The Empirical Rule is a key principle related to normally distributed datasets. It provides a quick way to understand the variability of data in itself and how much of the data falls within certain intervals. According to this rule:
  • Approximately 68% of the data falls within one standard deviation of the mean.
  • About 95% falls within two standard deviations.
  • Almost 99.7% falls within three standard deviations.

In the context of our farm workers' income study, we use the Empirical Rule to predict that nearly all sample means will fall within three standard deviations of the true population mean. Here, with a calculated standard deviation of 16 for the sample mean (from our large sample size calculation), this means the sample mean will fall within \(500 \pm 48\) New Zealand dollars, thanks to this rule.
By using the Empirical Rule, we can support the idea that observing a sample mean within \(\pm48\) of the population mean is almost certain. This simplifies understanding how variable or consistent our sample mean might be, giving us confidence in its reliability.
Standard Deviation
Standard deviation is a critical measure in statistics, indicating how spread out the data points are from the mean. In the case of a normal distribution, it tells us how much individual data points differ from the average.
For the farm workers' income distribution, the given standard deviation is \(\sigma = 160\) dollars. When dealing with a sample mean, like in this example, we adjust the original standard deviation based on the sample size using the formula:
\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]

For a sample of 100 workers, this leads to a standard deviation of \(\sigma_{\bar{x}} = 16\). This lower value for the sample means reflects less variability in the average of the sample compared to individual values, indicating how more data results in a more stable average.
Understanding standard deviation in this context helps clarify why our predictions about the sample mean using the Empirical Rule and CLT remain reliable. It emphasizes the sample's statistical stability, rendering our estimates both practical and interpretable.

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Most popular questions from this chapter

Aunt Erma's restaurant In Example 5 about Aunt Erma's Restaurant, the daily sales follow a probability distribution that has a mean of \(\mu=\$ 900\) and a standard deviation of \(\sigma=\$ 300\). This past week the daily sales for the seven days had a mean of \(\$ 980\) and a standard deviation of \(\$ 276\). a. Identify the mean and standard deviation of the population distribution. b. Identify the mean and standard deviation of the data distribution. What does the standard deviation describe? c. Identify the mean and the standard deviation of the sampling distribution of the sample mean for samples of seven daily sales. What does this standard deviation describe?

Finite populations The formula \(\sigma / \sqrt{n}\) for the standard deviation of \(\bar{x}\) actually is an approximation that treats the population size as infinitely large relative to the sample size \(n\). The exact formula for a finite population size \(N\) is Standard deviation \(=\sqrt{\frac{N-n}{N-1}} \frac{\sigma}{\sqrt{n}}\) The term \(\sqrt{(N-n) /(N-1)}\) is called the finite population correction. a. When \(n=300\) students are selected from a college student body of size \(\mathrm{N}=30,000\), show that the standard deviation equals \(0.995 \sigma / \sqrt{n}\). (When \(n\) is small compared to the population size \(\mathrm{N},\) the approximate formula works very well.) b. If \(n=\mathrm{N}\) (that is, we sample the entire population), show that the standard deviation equals \(0 .\) In other words, no sampling error occurs, since \(\bar{x}=\mu\) in that case.

Basketball shooting In college basketball, a shot made from beyond a designated arc radiating about 20 feet from the basket is worth three points, instead of the usual two points given for shots made inside that arc. Over his career, University of Florida basketball player Lee Humphrey made \(45 \%\) of his three-point attempts. In one game in his final season, he made only 3 of 12 three-point shots, leading a TV basketball analyst to announce that Humphrey was in a shooting slump. a. Assuming Humphrey has a \(45 \%\) chance of making any particular three-point shot, find the mean and standard deviation of the sampling distribution of the proportion of three-point shots he will make out of 12 shots. b. How many standard deviations from the mean is this game's result of making 3 of 12 three-point shots? c. If Humphrey was actually not in a slump but still had a \(45 \%\) chance of making any particular three-point shot, explain why it would not be especially surprising for him to make only 3 of 12 shots. Thus, this is not really evidence of a shooting slump.

What is a sampling distribution? How would you explain to someone who has never studied statistics what a sampling distribution is? Explain by using the example of polls of 1000 Canadians for estimating the proportion who think the prime minister is doing a good job.

Standard deviation of a proportion Suppose \(x=1\) with probability \(p,\) and \(x=0\) with probability \((1-p) .\) Then, \(x\) is the special case of a binomial random variable with \(n=1,\) so that \(\sigma=\sqrt{n p(1-p)}=\sqrt{p(1-p)} .\) With \(n\) trials, using the formula \(\sigma / \sqrt{n}\) for a standard deviation of a sample mean, explain why the standard deviation of a sample proportion equals \(\sqrt{p(1-p) / n}\)

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