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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 approximately (23.35, 25.65). c. The margin of error of the estimate is about 1.15.

Step by step solution

01

Find the point estimate of \(\mu\)

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

Construct a 99% confidence interval for \(\mu\)

This involves using the formula for a confidence interval, which is \(\bar{x} \pm Z(\frac{\sigma}{\sqrt{n}})\), where \(Z\) is the Z-score corresponding to the desired level of confidence. For a 99% confidence interval, the Z-score is approximately 2.57. Plugging in the given values, the confidence interval is 24.5 \pm 2.57(\frac{3.1}{\sqrt{20}}).
03

Calculate the margin of error

The margin of error is the range by which we allow the estimate to vary, it will form the length half the confidence interval. In this case, it is simply the product of the Z-score and standard error, which in this case comes out as 2.57*(\frac{3.1}{\sqrt{20}}).

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Key Concepts

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

Point Estimate
In statistics, a point estimate refers to using a single value as a best guess for an unknown population parameter, like the mean or proportion. When you're dealing with sample data, and you calculate an average, you're really trying to get a good sense of the overall average of the whole population. That sample mean (denoted as \(\bar{x}\)) is what we call a point estimate.

This is because it "points" us towards the population mean (\(\mu\)) directly using the information from the sample. In the given exercise, the point estimate for \(\mu\) is straightforwardly the sample mean, which is 24.5.

Key things to remember about point estimates:
  • They offer a single, specific number.
  • They're derived directly from the sample data.
  • Offer simplicity though lacking the precision of a range, which is where confidence intervals come into play.
Margin of Error
The margin of error gives insights into the precision of the point estimate. Think of it like a cushion around your estimate that tries to account for any potential errors that could arise simply because you are working with a sample and not the entire population.

In statistical terms, it is half the width of the confidence interval. It's basically adding and subtracting this cushion from the point estimate to specify a range where the true population parameter likely resides.

To calculate the margin of error, you multiply the Z-score, connected to your confidence level, by the standard error of the sample mean, \(\frac{\sigma}{\sqrt{n}}\). For instance, if you have a Z-score of 2.57 for a 99% confidence level, and your standard error is approximately 0.694, the margin of error comes out to be about 1.78.

Why is this handy?
  • Quantifies the uncertainty in your estimate.
  • Helps stakeholders understand the reliability of the estimate.
  • Essential to forming confidence intervals.
Z-Score
The Z-score is a crucial component when it comes to calculating confidence intervals. It's a statistical tool that measures how many standard deviations an element is from the mean. However, in this context, it's used to signify how confident we are about the estimates when constructing the interval around the point estimate.

When you determine your desired confidence level (such as 95% or 99%), the Z-score allows you to translate this into how precise the interval should be. For example, a 99% confidence level has a Z-score around 2.57, which means values up to 2.57 standard deviations away from the mean are considered.

This helps form the backbone of the confidence interval calculation:
  • Higher Z-scores mean a wider range, which implies more confidence but less precision.
  • They rely on the assumption of a normal distribution, especially important for smaller samples.
  • Essential for understanding the breadth of the interval and thus the reliability of the overall estimate.
Understanding Z-scores helps in demystifying confidence intervals and making statistically informed decisions.

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Most popular questions from this chapter

For a population, the value of the standard deviation is 4.96. A sample of 32 observations taken from this population produced the following data. \(\begin{array}{llllllll}74 & 85 & 72 & 73 & 86 & 81 & 77 & 60 \\ 83 & 78 & 79 & 88 & 76 & 73 & 84 & 78 \\ 81 & 72 & 82 & 81 & 79 & 83 & 88 & 86 \\ 78 & 83 & 87 & 82 & 80 & 84 & 76 & 74\end{array}\) 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?

What assumptions must hold true to use the \(t\) distribution to make a confidence interval for \(\mu\) ?

At the end of Section \(8.2\), 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 random sample of 16 airline passengers at the Bay City airport showed that the mean time spent waiting in line to check in at the ticket counters was 31 minutes with a standard deviation of 7 minutes. Construct a \(99 \%\) confidence interval for the mean time spent waiting in line by all passengers at this airport. Assume that such waiting times for all passengers are normally distributed.

a. A sample of 400 observations taken from a population produced a sample mean equal to \(92.45\) and a standard deviation equal to \(12.20 .\) Make a \(98 \%\) confidence interval for \(\mu\). b. Another sample of 400 observations taken from the same population produced a sample mean equal to \(91.75\) and a standard deviation equal to \(14.50 .\) Make a \(98 \%\) confidence interval for \(\mu .\) c. A third sample of 400 observations taken from the same population produced a sample mean equal to \(89.63\) and a standard deviation equal to \(13.40 .\) Make a \(98 \%\) confidence interval for \(\mu\). d. The true population mean for this population is \(90.65 .\) Which of the confidence intervals constructed in parts a through c cover this population mean and which do not?

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