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Chapter 8: Problem 46

A sample of 18 observations taken from a normally distributed population produced the following data: \(\begin{array}{lllllllll}28.4 & 27.3 & 25.5 & 25.5 & 31.1 & 23.0 & 26.3 & 24.6 & 28.4\end{array}\) \(\begin{array}{llllllllll}37.2 & 23.9 & 28.7 & 27.9 & 25.1 & 27.2 & 25.3 & 22.6 & 22.7\end{array}\) 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 \(\mu\) in part b?

Short Answer

Expert verified
The point estimate of \(\mu\) is the sample mean. The 99% confidence interval for \(\mu\) is calculated using the formula \(\overline{X} \pm z \cdot (s/\sqrt{n})\). The margin of error is \(z \cdot (s/\sqrt{n})\).

Step by step solution

01

Calculating the Sample Mean

First we add up all the observations and divide the sum by the total number of observations (n=18) to get the sample mean (\(\overline{X}\)).
02

Calculating the Sample Standard Deviation

Next, we subtract each observation from the mean, square the result, add up these squared values, divide by the number of observations minus 1 (n-1), and finally take the square root to get the standard deviation (\(s\)).
03

Point Estimate of \(\mu\)

The point estimate of \(\mu\) is simply the sample mean (\(\overline{X}\)), which we already calculated in Step 1.
04

Constructing a 99% Confidence Interval for \(\mu\)

To do this, we'll need to get the value of z that corresponds to a 99% confidence level (z = 2.575). The formula to calculate the confidence interval is \(\overline{X} \pm z \cdot (s/\sqrt{n})\).
05

Calculating the Margin of Error

The margin of error is simply the value we add and subtract from the sample mean to get the confidence interval. In this case, it's \(z \cdot (s/\sqrt{n})\).

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Key Concepts

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

Point Estimate
In statistical estimation, the point estimate is a single value that serves as an approximate value of an unknown parameter. For a population mean (\(\mu\)), the best point estimate is the sample mean (\(\overline{X}\)).
  • The point estimate is considered a good starting point for understanding the population mean.
  • However, it is important to note that a point estimate does not provide information about the potential error.
In the original exercise, determining the point estimate involved calculating the sample mean of the given observations. By summing all the data points and dividing by the total number of observations (which is 18), you arrive at the sample mean, which acts as the point estimate of the population mean.
Confidence Interval
A confidence interval provides a range of values that is likely to contain the population parameter, giving insights into the variability and reliability of the point estimate.
  • The confidence level (e.g., 99%) indicates how certain we are that the interval contains the true population parameter.
  • The wider the interval, the less precise, but potentially more reliable, the estimate.
  • Conversely, a narrower interval is more precise but has a higher risk of excluding the true mean.
Constructing a confidence interval in the exercise involved calculating it based on the sample mean and the sample standard deviation, incorporating the z-value associated with the desired confidence level (99% in this case) to account for sampling variability.
Sample Mean
The sample mean (\(\overline{X}\)) is computed by summing all of the data points in a sample and dividing by the total number of points.
  • The sample mean provides a simple, yet powerful, method to summarize data.
  • It is a critical component in the calculation of both the confidence interval and the margin of error.
The exercise demonstrates how to find the sample mean to serve as the point estimate for the population's mean. By working with the sample data, we are able to derive an average that estimates the central tendency of the entire population.
Margin of Error
The margin of error quantifies the uncertainty in the point estimate. It provides a buffer zone around the sample mean to account for random sampling errors and can greatly influence the width of a confidence interval.
  • The formula used to calculate the margin of error is \( z \cdot \left( \frac{s}{\sqrt{n}} \right) \), where \(z\) is the z-score related to the chosen confidence level, \(s\) is the sample standard deviation, and \(n\) is the sample size.
  • A larger margin of error gives a wider range for the confidence interval, ensuring higher reliability that it contains the population mean.
  • Choosing a higher confidence level will increase the margin of error, illustrating the trade-off between precision and confidence.
In the example from the exercise, the margin of error for the 99% confidence interval was calculated by multiplying the z-score with the standard error, thereby illustrating how much the sample mean could vary from the population mean.

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Most popular questions from this chapter

a. How large a sample should be selected so that the margin of error of estimate for a \(98 \%\) confidence interval for \(p\) is \(.045\) when the value of the sample proportion obtained from a preliminary sample is \(.53\) ? b. Find the most conservative sample size that will produce the margin of error for a \(98 \%\) confidence interval for \(p\) equal to \(.045\)

A survey of 500 randomly selected adult men showed that the mean time they spend per week watching sports on television is \(9.75\) hours with a standard deviation of \(2.2\) hours. Construct a \(90 \%\) confidence interval for the population mean, \(\mu .\)

A random sample of 36 mid-sized cars tested for fuel consumption gave a mean of \(26.4\) miles per gallon with a standard deviation of \(2.3\) miles per gallon. a. Find a \(99 \%\) confidence interval for the population mean, \(\mu .\) b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Describe all possible alternatives. Which alternative is the best and why?

A department store manager wants to estimate at a \(98 \%\) confidence level the mean amount spent by all customers at this store. The manager knows that the standard deviation of amounts spent by all customers at this store is \(\$ 31\). What sample size should he choose so that the estimate is within \(\$ 3\) of the population mean?

Salaried workers at a large corporation receive 2 weeks' paid vacation per year. Sixteen randomly selected workers from this corporation were asked whether or not they would be willing to take a \(3 \%\) reduction in their annual salaries in return for 2 additional weeks of paid vacation. The following are the responses of these workers. \(\begin{array}{llllllll}\text { No } & \text { Yes } & \text { No } & \text { No } & \text { Yes } & \text { No } & \text { No } & \text { Yes } \\ \text { Yes } & \text { No } & \text { No } & \text { No } & \text { Yes } & \text { No } & \text { No } & \text { No }\end{array}\) Construct a \(97 \%\) confidence interval for the percentage of all salaried workers at this corporation who would accept a \(3 \%\) pay cut in return for 2 additional weeks of paid vacation.
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