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Answer true or false. Explain your answer. The point estimate for the population mean \(\mu\) of an \(x\) distribution is \(\bar{x},\) computed from a random sample of the \(x\) distribution.

Short Answer

Expert verified
True. The sample mean \(\bar{x}\) is the point estimate for the population mean \(\mu\).

Step by step solution

01

Understanding Point Estimates

Recall that a point estimate is a single value given as an estimate of a population parameter. For the population mean \(\mu\), the sample mean \(\bar{x}\) is the most common point estimate.
02

Defining Terms

The population mean \(\mu\) is the average of all observations in the entire population, while the sample mean \(\bar{x}\) is the average of the observations in a sample taken from that population.
03

Point Estimate for Population Mean

In statistics, the sample mean \(\bar{x}\) is used as a point estimate of the population mean \(\mu\). This is because it provides the best unbiased estimate based on the sample data.
04

Conclusion

Since \(\bar{x}\) is used as a point estimate for \(\mu\) in a random sample drawn from the population, the statement is indeed true.

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Key Concepts

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

Understanding the Population Mean
The population mean, denoted by the symbol \(\mu\), represents the average of all values in a population. It is an essential parameter in statistics because it summarizes a central tendency of the entire group. Think of it as the balancing point of a seesaw. Imagine you have a large jar filled with countless candies. If you wanted to know the average weight of all the candies in the jar, you would find the population mean. Calculating the population mean involves summing up all the individual observations in the population and dividing by the total number of observations. In practice, finding the population mean can be difficult when dealing with large populations, as it's often impossible or impractical to inspect every member. This is where sample means come into play.
Sample Mean as a Reliable Point Estimate
The sample mean, represented by \(\bar{x}\), is the average of observations obtained from a sample drawn from a population. It's calculated exactly like the population mean but only involves the values in the sample, not the entire population. Let's go back to our jar of candies. Instead of weighing every candy, you take a handful, weigh them, and then calculate their average weight. This average would be your sample mean. The sample mean is a crucial tool in statistics because it acts as the point estimate for the population mean \(\mu\). One of the strengths of the sample mean is its efficiency.
  • It's straightforward to calculate.
  • It provides a good approximation of the population mean if the sample is random and representative.
The sample mean can closely approximate the population mean even if the sample size is small, thanks to the central limit theorem.
The Concept of Unbiased Estimate
An unbiased estimate is one where the expected value of the estimate equals the true value of the parameter being estimated. In statistics, unbiasedness is a desirable property because it means on average, the estimate reflects the true parameter value without systematic error.When we say the sample mean \(\bar{x}\) is an unbiased estimate of the population mean \(\mu\), it means that if you were to draw many samples and compute their means, the average of these sample means would equal the true population mean. It's like practicing free throws in basketball. If you shoot enough times, the average number of successful shots will represent your true skill level.
  • The sample mean is unbiased because it doesn't consistently overestimate or underestimate \(\mu\).
  • This characteristic makes the sample mean a trusted tool for estimating population parameters in statistics.
When using the sample mean as a point estimate, no systematic errors are introduced, providing a reliable measure of the population mean.

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Most popular questions from this chapter

Independent random samples of professional football and basketball players gave the following information (References: Sports Encyclopedia of Pro Football and Official NBA Basketball Encyclopedia ). Note: These data are also available for download at the Companion Sites for this text. Assume that the weight distributions are mound-shaped and symmetric. (a) Use a calculator with mean and standard deviation keys to verify that \(\bar{x}_{1}=259.6, s_{1}=12.1, \bar{x}_{2}=205.8,\) and \(s_{2}=12.9\) (b) Let \(\mu_{1}\) be the population mean for \(x_{1}\) and let \(\mu_{2}\) be the population mean for \(x_{2} .\) Find a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) (c) Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the \(99 \%\) level of confidence, do professional football players tend to have a higher population mean weight than professional basketball players? (d) Which distribution (standard normal or Student's \(t\) ) did you use? Why?

Lorraine was in a hurry when she computed a confidence interval for \(\mu\). Because \(\sigma\) was not known, she used a Student's \(t\) distribution. However, she accidentally used degrees of freedom \(n\) instead of \(n-1 .\) Was her confidence interval longer or shorter than one found using the correct degrees of freedom \(n-1 ?\) Explain.

A New York Times/CBS poll asked the question, "What do you think is the most important problem facing this country today?" Nineteen percent of the respondents answered, "Crime and violence." The margin of sampling error was plus or minus 3 percentage points. Following the convention that the margin of error is based on a \(95 \%\) confidence interval, find a \(95 \%\) confidence interval for the percentage of the population that would respond, "Crime and violence" to the question asked by the pollsters.

In a survey of 1000 large corporations, 250 said that, given a choice between a job candidate who smokes and an equally qualified nonsmoker, the nonsmoker would get the job (USA Today). (a) Let \(p\) represent the proportion of all corporations preferring a nonsmoking candidate. Find a point estimate for \(p\). (b) Find a 0.95 confidence interval for \(p\). (c) As a news writer, how would you report the survey results regarding the proportion of corporations that hire the equally qualified nonsmoker? What is the margin of error based on a \(95 \%\) confidence interval?

Santa Fe black-on-white is a type of pottery commonly found at archacological excavations in Bandelier National Monument. At one excavation site a sample of 592 potsherds was found, of which 360 were identified as Santa Fe black-on- white (Bandelier Archaeological Excavation Project: Summer 1990 Excavations at Burnt Mesa Pueblo and Casa del Rito, edited by Kohler and Root, Washington State University). (a) Let \(p\) represent the population proportion of Santa Fe black-on-white potsherds at the excavation site. Find a point estimate for \(p\) (b) Find a \(95 \%\) confidence interval for \(p\). Give a bricf statement of the meaning of the confidence interval. (c) Do you think the conditions \(n p > 5\) and \(n q > 5\) are satisfied in this problem? Why would this be important?

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