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Chapter 9: Problem 32

The American Restaurant Association collected information on the number of meals eaten outside the home per week by young married couples. A survey of 60 couples showed the sample mean number of meals eaten outside the home was 2.76 meals per week, with a standard deviation of 0.75 meal per week. Construct a \(99 \%\) confidence interval for the population mean.

Short Answer

Expert verified
The 99% confidence interval is approximately (2.51, 3.01).

Step by step solution

01

Identify the Required Information

To construct a confidence interval, we need the sample mean \( \bar{x} = 2.76 \), the sample standard deviation \( s = 0.75 \), the sample size \( n = 60 \), and the confidence level \( 99\% \).
02

Find the Critical Value

Since the sample size is 60, which is greater than 30, we use the standard normal distribution (Z-distribution). For a \(99\%\) confidence interval, the critical value \( z \) is approximately 2.576.
03

Calculate the Standard Error

The standard error (SE) of the mean is given by the formula: \[ SE = \frac{s}{\sqrt{n}} = \frac{0.75}{\sqrt{60}} \approx 0.0968 \]
04

Construct the Confidence Interval

The confidence interval is calculated using the formula: \[ \bar{x} \pm z \times SE \] Substituting the known values: \[ 2.76 \pm 2.576 \times 0.0968 \] This results in: \[ 2.76 \pm 0.2491 \] Calculating the bounds, we find: Lower bound = 2.76 - 0.2491 = 2.5109, Upper bound = 2.76 + 0.2491 = 3.0091.

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Key Concepts

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

Sample Mean
In statistics, the sample mean is a measure that represents the average of a set of data points collected from a sample. It is denoted as \( \bar{x} \).
  • Finding the sample mean involves summing up all the values in the dataset and then dividing by the number of observations in the sample.
  • The sample mean is used to make inferences about the population mean, which is the average of an entire population.
For example, in the problem where we are analyzing young married couples' eating habits, the sample mean number of meals eaten outside the home was calculated to be 2.76 meals per week.
Standard Deviation
The standard deviation is a key statistical measure that gives insight into the variability or dispersion of a dataset. It is denoted by \( s \) for a sample.
  • It tells us how spread out the data points are around the mean. A higher standard deviation indicates more spread, while a lower standard deviation suggests that the data is clustered closely around the mean.
  • Standard deviation is crucial when assessing how typical a particular data point is within the group.
In our example, the standard deviation of 0.75 indicates how much the individual numbers of meals eaten outside the home vary from the sample mean of 2.76.
Standard Error
The standard error (SE) is an important concept in inferential statistics, as it provides a sense of how much the sample mean might differ from the true population mean. It is calculated by dividing the standard deviation \( s \) by the square root of the sample size \( n \): \[ SE = \frac{s}{\sqrt{n}} \]
  • The standard error decreases as the sample size increases, reflecting more precise estimates of the population mean with larger samples.
  • In confidence interval calculations, the standard error gauges the uncertainty around the sample mean estimate.
In our case, with a standard deviation of 0.75 and a sample size of 60, the standard error was calculated to be approximately 0.0968.
Z-distribution
The Z-distribution, also known as the standard normal distribution, is a fundamental concept in statistics used for comparing scores from different normal distributions or deriving inferences.
  • It is symmetrical and bell-shaped, with a mean of 0 and a standard deviation of 1.
  • The Z-distribution is used to find the critical values necessary to construct confidence intervals when sample sizes are large (typically, \( n \geq 30 \)).
  • In our problem, the critical value of 2.576 was chosen because we are constructing a 99% confidence interval, which requires us to capture 99% of all possible sample means around our known sample mean.
This choice helps in effectively estimating the range within which the true population mean may lie based on our sample data.

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Most popular questions from this chapter

An important factor in selling a residential property is the number of times real estate agents show a home. A sample of 15 homes recently sold in the Buffalo, New York, area revealed the mean number of times a home was shown was 24 and the standard deviation of the sample was 5 people. Develop a \(98 \%\) confidence interval for the population mean.

You plan to conduct a survey to find what proportion of the workforce has two or more jobs. You decide on the \(95 \%\) confidence level and a margin of error of \(2 \% .\) A pilot survey reveals that 5 of the 50 sampled hold two or more jobs. How many in the workforce should be interviewed to meet your requirements?

As part of an annual review of its accounts, a discount brokerage selected a random sample of 36 customers and reviewed the value of their accounts. The mean was 32,000 with a sample standard deviation of 8,200 . What is a \(90 \%\) confidence interval for the mean account value of the population of customers?

Ms. Maria Wilson is considering running for mayor of Bear Gulch, Montana. Before completing the petitions, she decides to conduct a survey of voters in Bear Gulch. A sample of 400 voters reveals that 300 would support her in the November election. a. Estimate the value of the population proportion. b. Develop a \(99 \%\) confidence interval for the population proportion. c. Interpret your findings.

A recent study by the American Automobile Dealers Association surveyed a random sample of 20 dealers. The data revealed a mean amount of profit per car sold was 290, with a standard deviation of 125 . Develop a \(95 \%\) confidence interval for the population mean of profit per car.
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