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Chapter 8: Problem 106

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, \(\mu .\)

Short Answer

Expert verified
The 90% confidence interval for the population mean is \(9.75 \pm 1.645 \cdot \frac{2.2}{\sqrt{500}}\)

Step by step solution

01

Compute Standard Error

First, we need to compute the standard error (SE) of the mean. The standard error is calculated as \(SE = \frac{s}{\sqrt{n}}\) where \(s\) is the standard deviation and \(n\) is the sample size. Substituting the respective values, we have \(SE = \frac{2.2}{\sqrt{500}}\).
02

Find z-score

To construct the 90% confidence interval, we need to find the z-score associated with the middle 90% of the data. From the standard normal distribution, the z-score that cuts off the bottom 5% and the top 5% (to leave the middle 90%) is approximately \(\pm 1.645\).
03

Construct Confidence Interval

The confidence interval is then constructed as follows: \(\bar{x} \pm z \cdot SE\), where \(\bar{x}\) is the sample mean and \(z\) is the z-score for the desired level of confidence. Now, substituting the respective values, we get: \(9.75 \pm 1.645 \cdot SE\).
04

Compute Confidence Interval

Finally, calculate the confidence interval by substituting the value of SE computed in Step 1 into the formula from Step 3. Multiply the z-score by the standard error and add this product to and subtract it from the sample mean to get the confidence range.

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Key Concepts

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

Standard Error
When working with statistical data, we often deal with samples, not the entire population. This means there is some variability in the data we collect. To measure this variability, we use the Standard Error (SE) of the mean. It tells us how much the sample mean is expected to deviate from the true population mean.The formula for the standard error is given by:\[ SE = \frac{s}{\sqrt{n}} \]
  • **Where**:
    • \(s\) is the standard deviation of the sample
    • \(n\) is the size of the sample.
The smaller the standard error, the more accurate our sample mean is likely to be an estimation of the population mean.
Conversely, a larger standard error indicates more variability and less certainty about the sample mean as a representation of the population mean.
Z-score
The Z-score is a crucial component in statistics, especially when constructing confidence intervals. It represents the number of standard deviations a data point is from the mean.For confidence intervals, the Z-score helps us understand how confident we can be that our sample mean reflects the true population mean.Depending on the confidence level desired (e.g., 90%, 95%, 99%), the Z-score will be different:
  • For 90% confidence, the Z-score is approximately \( \pm 1.645 \)
  • For 95% confidence, it is approximately \( \pm 1.96 \)
  • For 99% confidence, it's around \( \pm 2.576 \)
This Z-score is used in calculating the range within which the true population mean is likely to fall.
By incorporating the Z-score into the confidence interval formula, we determine the margin of error around the sample mean.
Population Mean
The population mean, denoted as \( \mu \), is the average of all measurements in a population. In practice, it is difficult to measure every individual in a large population. Thus, we rely on sample data to estimate the population mean. When constructing a confidence interval, our goal is to make an educated guess about this population mean using the sample mean and some statistical calculations.This estimation of the population mean is never exact, but using statistics, we are able to estimate a range of values (confidence interval) that we are reasonably sure \( \mu \) lies within.
This allows us to make better-informed decisions based on sample data.
Sample Mean
The sample mean, represented by \( \bar{x} \), is simply the average value of the data points in a sample drawn from a population.Calculating the sample mean is straightforward. The formula is:\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \]
  • **Where** \( x_i \) represents each data point in the sample, and \( n \) is the number of data points in the sample.
The sample mean serves as our best estimate of the population mean. Therefore, it plays a key role in constructing confidence intervals.A well-calculated sample mean, alongside other statistical measures, guides us in understanding the population without surveying every member.
This statistical approach saves time and resources while providing valuable insights.

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Most popular questions from this chapter

A marketing researcher wants to find a \(95 \%\) confidence interval for the mean amount that visitors to a theme park spend per person per day. She knows that the standard deviation of the amounts spent per person per day by all visitors to this park is \(\$ 11\). How large a sample should the researcher select so that the estimate will be within \(\$ 2\) of the population mean?

A local gasoline dealership in a small town wants to estimate the average amount of gasoline that people in that town use in a 1-week period. The dealer asked 44 randomly selected customers to keep a diary of their gasoline usage, and this information produced the following data on gas used (in gallons) by these people during a 1-week period. The population standard deviation is not known. \(\begin{array}{rrrrrrrrrrr}23.1 & 13.6 & 25.8 & 10.0 & 7.6 & 18.9 & 26.6 & 23.8 & 12.3 & 15.8 & 21.0 \\ 26.9 & 22.9 & 18.3 & 23.5 & 21.6 & 15.5 & 23.5 & 11.8 & 15.3 & 11.9 & 19.2 \\ 14.5 & 9.6 & 12.1 & 18.0 & 20.6 & 14.2 & 7.1 & 13.2 & 5.3 & 13.1 & 10.9 \\ 10.5 & 5.1 & 5.2 & 6.5 & 8.3 & 10.5 & 7.4 & 7.4 & 5.3 & 10.6 & 13.0\end{array}\) Construct a \(95 \%\) confidence interval for the average weekly gas usage by people in this town.

A company that produces detergents wants to estimate the mean amount of detergent in 64 -ounce jugs at a \(99 \%\) confidence level. The company knows that the standard deviation of the amounts of detergent in all such jugs is \(.20\) ounce. How large a sample should the company select so that the estimate is within \(.04\) ounce of the population mean?

Because of inadequate public school budgets and lack of money available to teachers for classroom materials, many teachers often use their own money to buy materials used in the classrooms. A random sample of 100 public school teachers selected from an eastern state showed that they spent an average of \(\$ 290\) of their own money on such materials during the 2011-2012 school year. The population standard deviation was \(\$ 70\). a. What is the point estimate of the mean of such expenses incurred during the \(2011-2012\) school year by all public school teachers in this state? b. Make a \(95 \%\) confidence interval for the corresponding population mean.

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.
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