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Chapter 8: Problem 3

Explain the difference between a point estimate and an interval estimate.

Short Answer

Expert verified
A point estimate provides a single, specific estimate of an unknown parameter while an interval estimate provides a range of estimates for an unknown parameter.

Step by step solution

01

Define Point Estimate

A Point Estimate is a statistical strategy where you ascertain just one approximation to guess the value of an unknown parameter. It provides a single guess of what the parameter could be. For instance, an average might be used to anticipate the exact output. These estimations can be potentially erroneous as they set the possible value to just one point and other variables are not considered in the estimation.
02

Define Interval Estimate

An Interval estimate, instead of being a single value (like a point estimate), it gives a range of probable values for the unknown parameter. It takes into account variability. The interval estimate provides a range within which we can say with a certain level of confidence that the unknown population parameter lies.
03

Distinguish between the two

The main difference between an interval estimate and a point estimate lies in the fact that a point estimate is a single value while an interval estimate is a range of values. A point estimate does not account for variability, hence can be potentially erroneous while an interval estimate gives a range of values reflecting the potential variability and the level of uncertainty adding to its accuracy.

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Key Concepts

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

Point Estimate
A point estimate provides a single best guess of an unknown parameter. It tries to identify the exact value of a population parameter by deriving an estimate from a sample statistic. For example, the sample mean might be used as a point estimate of the population mean.
This method is straightforward but can be inaccurate. Why? Because if the estimated point does not capture the true value, errors are inevitable. Variability is not considered, leading to a higher chance of divergence between the point estimate and the true value.
Nevertheless, point estimates are a useful starting point for statistical analyses, offering a concise, albeit potentially limited, estimation of an unknown parameter.
Interval Estimate
An interval estimate extends the idea of a point estimate by providing a range of values within which the unknown parameter is likely to lie. This range is determined by including a measure of variability, thereby increasing the estimate's reliability.
Typically, interval estimates are expressed with a confidence interval. For instance, "We are 95% confident that the true population mean falls within this range." This inclusion of uncertainty acknowledges real-world complexities beyond simple point estimation.
Factors such as sample size and variability affect the width of the confidence interval. Larger samples generally result in narrower intervals, reflecting greater precision in estimating the unknown parameter.
Unknown Parameter
In statistics, an unknown parameter represents the true yet unknown value of a quantity within a population. This could be the mean, variance, or any other characteristic of the population that we seek to uncover through statistical estimation.
The goal of both point and interval estimates is to infer this unknown parameter based on sample data.
While the parameter is inherently unknown, estimates help in making educated guesses. Understanding the limitations and strengths of both point and interval estimates is vital for accurate statistical analysis, painting a clearer picture of what the parameter might actually be.

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Most popular questions from this chapter

Suppose that a confidence interval is assigned a level of confidence of \(1-\alpha=95 \% .\) How is the \(95 \%\) used in constructing the confidence interval? If \(1-\alpha\) was changed to \(90 \%,\) what effect would this have on the confidence interval?

Use a computer or calculator to select 40 random single-digit numbers. Find the sample mean, \(z \star,\) and \(p\) -value for testing \(H_{o}: \mu=4.5\) against a two-tailed alternative. Repeat several times as in Table \(8.8 .\) Describe your findings.

A lawn and garden sprinkler system is designed to have a delayed start; that is, there is a delay from the moment it is turned on until the water starts. The delay times form a normal distribution with mean 45 seconds and standard deviation 8 seconds. Several customers have complained that the delay time is considerably longer than claimed. The system engineer has selected a random sample of 15 installed systems and has obtained one delay time from each system. The sample mean is 50.1 seconds. Using \(\alpha=0.02,\) is there significant evidence to show that the customers might be correct that the mean delay time is more than 45 seconds? a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

Women own an average of 15 pairs of shoes. This is based on a survey of female adults by Kelton Research for Eneslow, the New York City-based Foot Comfort Center. Suppose a random sample of 35 newly hired female college graduates was taken and the sample mean was 18.37 pairs of shoes. If \(\sigma=6.12,\) does this sample provide sufficient evidence that young female college graduates' mean number of shoes is greater than the overall mean number for all female adults? Use a 0.10 level of significance.

Assume that \(z\) is the test statistic and calculate the value of \(z \star\) for each of the following: a. \(\quad H_{o}: \mu=51, \sigma=4.5, n=40, \bar{x}=49.6\) b. \(\quad H_{o}: \mu=20, \sigma=4.3, n=75, \bar{x}=21.2\) c. \(\quad H_{o}: \mu=138.5, \sigma=3.7, n=14, \bar{x}=142.93\) d. \(\quad H_{o}: \mu=815, \sigma=43.3, n=60, \bar{x}=799.6\)
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