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Chapter 8: Problem 11

For a data set obtained from a sample, \(n=20\) and \(\bar{x}=24.5 .\) It is known that \(\sigma=3.1\). The population is normally distributed. a. What is the point estimate of \(\mu ?\) b. Make a \(99 \%\) confidence interval for \(\mu\). c. What is the margin of error of estimate for part b?

Short Answer

Expert verified
a. The point estimate of \(\mu\) is \(24.5\). b. The 99% confidence interval for \(\mu\) is \((22.71, 26.29)\). c. The margin of error for the confidence interval is \(1.79\).

Step by step solution

01

Calculate Point Estimate of Population Mean (\(\mu\))

The point estimate of the population mean (\(\mu\)) is simply given by the sample mean (\(\bar{x}\)). So, the point estimate of \(\mu\) is \(24.5\).
02

Determine the Appropriate Z-Score for a 99% Confidence Interval

A 99% confidence interval corresponds to an alpha level (\(\alpha\)) of \(0.01\) (i.e., \(100%-99% = 1% \rightarrow 0.01\)) divided by 2 (for a two-tailed test), giving \(\alpha/2 = 0.005\). Look up this value in a standard normal (z-score) table, and find the corresponding z-score, which is about \(2.576\).
03

Calculate the Margin of Error

For a z-score distribution, the margin of error (E) in a confidence interval is calculated using the formula: \(E = z \times \(\frac{\sigma}{\sqrt{n}}\)\). Substitute the known values into the formula to get \(E = 2.576 \times \(\frac{3.1}{\sqrt{20}}\) = 1.79\)
04

Construct the Confidence Interval

A confidence interval is determined by taking the point estimate and then adding and subtracting the margin of error. So, the confidence interval is \(24.5 卤 1.79 = (22.71, 26.29)\).

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Key Concepts

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

Point Estimate
The point estimate is like a quick snapshot of the overall picture when working with data from a sample. It is used to best approximate an unknown population parameter. In this context, the parameter we are trying to estimate is the population mean, denoted as \(\mu\). The simplest approach to find this value is by using the sample mean, represented as \(\bar{x}\).
Basically, the point estimate provides us with our best guess for \(\mu\). Since the sample mean for the given data set is 24.5, that will be our point estimate for \(\mu\). It gives us a foundational start for further analysis whenever deriving insights from sample data.
Sample Mean
The sample mean, represented by \(\bar{x}\), is crucial when working with data. It provides the average value of the data set you've collected. Gathering a sample mean gives a more manageable number to work with than tackling each data point individually.
  • This mean is calculated by summing up all data values in the sample and dividing the total by the number of data points.
  • It serves as the center or "balance point" of the data collected.

In the original exercise, the sample mean comes out to be 24.5. This number acts as our initial estimator for the true population mean, \(\mu\), providing us with insight into the general trend of the larger population from which this sample was drawn.
Margin of Error
The margin of error essentially tells us how much uncertainty we have in our point estimate. It provides a range on either side of the sample mean, creating a bandwidth in which the true population mean may lie with a certain level of confidence.
This is calculated by using the formula:
  • First, find the z-score associated with the desired confidence level.
  • Multiply this z-score by the standard deviation divided by the square root of the sample size: \(E = z \times \left(\frac{\sigma}{\sqrt{n}}\right)\).
The margin of error increases our understanding of the potential inaccuracy and sets realistic expectations.
In the exercise provided, using a 99% confidence interval and a z-score of approximately 2.576, we computed the margin of error to be approximately 1.79. This means our estimate is likely to be within 1.79 units of the sample mean.
Z-Score
A Z-Score helps indicate how much a statistic deviates from the mean, providing a way to understand where a data point sits within a distribution. In confidence intervals, the z-score allows us to generate a realistic range where we expect our population parameter (like \(\mu\)) to lie.
  • The value is given by a standard normal distribution table and relates to the chosen level of confidence.
  • For a 99% confidence level, the corresponding z-score is typically around 2.576, which is derived from having 0.5% (or 0.005) in each tail of the normal curve, given it's a two-tailed test.

This z-score is crucial as it expands the margin of error, broadening the confidence interval, and increasing our certainty that \(\mu\) is covered by this range. It bridges the know-how from statistics to practical real-world data evaluations.

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Most popular questions from this chapter

KidPix Entertainment is in the planning stages of producing a new computer- animated movie for national release, so they need to determine the production time (labor-hours necessary) to produce the movie. The mean production time for a random sample of 14 big-screen computer-animated movies is found to be 53,550 labor-hours. Suppose that the population standard deviation is known to be 7462 labor-hours and the distribution of production times is normal. a. Construct a \(98 \%\) confidence interval for the mean production time to produce a big-screen computer-animated movie. b. Explain why we need to make the confidence interval. Why is it not correct to say that the average production time needed to produce all big-screen computer-animated movies is 53,550 labor-hours?

Briefly explain the meaning of the degrees of freedom for a \(t\) distribution. Give one example.

At the end of Section \(8.3\), we noted that we always round up when calculating the minimum sample size for a confidence interval for \(\mu\) with a specified margin of error and confidence level. Using the formula for the margin of error, explain why we must always round up in this situation.

A sample of 11 observations taken from a normally distributed population produced the following data. \(\begin{array}{lllllllllll}-7.1 & 10.3 & 8.7 & -3.6 & -6.0 & -7.5 & 5.2 & 3.7 & 9.8 & -4.4 & 6.4\end{array}\) a. What is the point estimate of \(\mu\) ? b. Make a \(95 \%\) confidence interval for \(\mu\). c. What is the margin of error of estimate for \(\mu\) in part b?

\(8.104\) A random sample of 25 life insurance policyholders showed that the average premium they pay on their life insurance policies is \(\$ 685\) per year with a standard deviation of \(\$ 74\). Assuming that the life insurance policy premiums for all life insurance policyholders have a normal distribution, make a \(99 \%\) confidence interval for the population mean, \(\mu\).
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