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Playbill magazine reported that the mean annual household income of its readers is \(\$ 119,155\) (Playbill, January 2006 ). Assume this estimate of the mean annual household income is based on a sample of 80 households, and based on past studies, the population standard deviation is known to be \(\sigma=\$ 30,000\) a. Develop a \(90 \%\) confidence interval estimate of the population mean. b. Develop a \(95 \%\) confidence interval estimate of the population mean. c. Develop a \(99 \%\) confidence interval estimate of the population mean. d. Discuss what happens to the width of the confidence interval as the confidence level is increased. Does this result seem reasonable? Explain.

Short Answer

Expert verified
a) (113,639.76, 124,670.24); b) (112,581.28, 125,728.72); c) (110,518.25, 127,791.75); d) Wider intervals occur with higher confidence.

Step by step solution

01

Understanding the Concept

A confidence interval provides a range of values that is likely to contain the population mean with a certain level of confidence. It is calculated using statistical formulas that incorporate the sample mean, the standard deviation of the population, the sample size, and a critical value from the standard normal distribution corresponding to the desired confidence level.
02

Gather Essential Data

We have a sample mean \(\bar{x} = 119,155\), population standard deviation \(\sigma = 30,000\), and sample size \(n = 80\). We will calculate three confidence intervals (90%, 95%, 99%) using these parameters.
03

Calculate the Standard Error

The standard error (SE) is calculated as the population standard deviation divided by the square root of the sample size: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{30,000}{\sqrt{80}} \approx 3353.94 \]
04

Determine Critical z-Values

For a 90% confidence interval, the critical z-value is approximately 1.645. For 95%, it is 1.96. For 99%, it is 2.576. These values are based on standard normal distribution tables.
05

Calculate Confidence Intervals

For each confidence level, the confidence interval is \[ \bar{x} \pm (z \times SE) \]- **90% Confidence Interval:** \[ 119,155 \pm (1.645 \times 3353.94) \approx 119,155 \pm 5515.24 \]\((113,639.76, 124,670.24)\) - **95% Confidence Interval:** \[ 119,155 \pm (1.96 \times 3353.94) \approx 119,155 \pm 6573.72 \]\((112,581.28, 125,728.72)\)- **99% Confidence Interval:** \[ 119,155 \pm (2.576 \times 3353.94) \approx 119,155 \pm 8636.75 \]\((110,518.25, 127,791.75)\)
06

Analyze the Effect of Confidence Level

As the confidence level increases from 90% to 99%, the width of the confidence interval also increases. This occurs because higher confidence levels require a wider range to ensure the population mean is captured within the interval. This is reasonable, as we trade off precision for confidence.

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Key Concepts

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

Population Mean
The concept of the population mean refers to the average of a group from which a sample is drawn. In statistics, it is often denoted by the symbol \( \mu \). It represents the central tendency of data and is a key component in determining confidence intervals.

When conducting surveys or studies, we typically cannot access the entire population, so we use the sample mean (denoted by \( \bar{x} \)) as an estimator of the population mean. In the exercise above, the sample mean was \( \$119,155\). This acts as our best estimate of the average income for the broader group of annual household incomes of Playbill readers.

Understanding the difference between the sample mean and the true population mean is crucial because the sample mean is only an approximation. Our confidence interval calculations aim to provide a range that likely includes the true mean.
Standard Error
The standard error measures the accuracy with which a sample mean represents the population mean. It gives insight into how much variation we can expect between the sample mean and the population mean if we were to draw multiple samples.

Typically, the standard error is calculated by dividing the population standard deviation (\( \sigma \)) by the square root of the sample size (\( n \)). In our example, the standard deviation is \( \$30,000\) and the sample size is \( 80 \). The formula becomes \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{30,000}{\sqrt{80}} \approx 3353.94 \]

A lower standard error indicates that the sample mean is a more accurate reflection of the population mean. The standard error is crucial when constructing confidence intervals because it tells us the "wiggle room" in our estimation.
Z-Values
Z-values are critical values derived from the standard normal distribution, used in calculating confidence intervals. They help determine how far, in standard deviations, a data point is from the mean.
  • For a 90% confidence interval, the critical z-value is 1.645.
  • For a 95% confidence interval, it is 1.96.
  • For a 99% confidence interval, it is 2.576.
These z-values are constant and are found in statistical tables. They represent the degree of certainty we require—higher confidence levels (like 99%) require a wider interval, hence a larger z-value.

The z-value is multiplied by the standard error to find the margin of error for the interval. The formula used is \[ \text{Margin of Error} = z \times SE \] Understanding z-values is essential because they form the backbone of how wide or narrow a confidence interval will be.
Confidence Level
The confidence level indicates the degree of certainty we have that a confidence interval includes the population mean. Higher confidence levels represent greater certainty but come at a cost—more considerable width in the confidence intervals.

Here's a breakdown:
  • At a 90% confidence level, there's a 90% chance that the interval contains the population mean.
  • A 95% confidence level offers more assurance, hence, the interval is wider.
  • At 99%, the interval is even wider, providing nearly certain containment of the mean.
Each level has its applications; for instance, tightly controlled experiments might lean towards 90% for narrower intervals, while exploratory research might use 95% or 99%. The trade-off is between precision and confidence—the higher the confidence, the wider the interval, because we need more room to "capture" the true mean. This is why as confidence increases (as demonstrated from 90% to 99% in the steps), the interval width also expands.

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Most popular questions from this chapter

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 65 weekly reports showed a sample mean of 19.5 customer contacts per week. The sample standard deviation was \(5.2 .\) Provide \(90 \%\) and \(95 \%\) confidence intervals for the population mean number of weekly customer contacts for the sales personnel.

A survey of small businesses with Web sites found that the average amount spent on a site was \(\$ 11,500\) per year (Fortune, March 5, 2001). Given a sample of 60 businesses and a population standard deviation of \(\sigma=\$ 4000\), what is the margin of error? Use \(95 \%\) confidence. What would you recommend if the study required a margin of error of \(\$ 500 ?\)

The average cost of a gallon of unleaded gasoline in Greater Cincinnati was reported to be \(\$ 2.41\) (The Cincinnati Enquirer, February 3,2006 ). During periods of rapidly changing prices, the newspaper samples service stations and prepares reports on gasoline prices frequently. Assume the standard deviation is \(\$ .15\) for the price of a gallon of unleaded regular gasoline, and recommend the appropriate sample size for the newspaper to use if they wish to report a margin of error at \(95 \%\) confidence. a. Suppose the desired margin of error is \(\$ .07\) b. Suppose the desired margin of error is \(\$ .05\) c. Suppose the desired margin of error is \(\$ .03\)

A simple random sample of 400 individuals provides 100 Yes responses. a. What is the point estimate of the proportion of the population that would provide Yes responses? b. What is your estimate of the standard error of the proportion, \(\sigma_{\bar{p}} ?\) c. Compute the \(95 \%\) confidence interval for the population proportion.

In developing patient appointment schedules, a medical center wants to estimate the mean time that a staff member spends with each patient. How large a sample should be taken if the desired margin of error is two minutes at a \(95 \%\) level of confidence? How large a sample should be taken for a \(99 \%\) level of confidence? Use a planning value for the population standard deviation of eight minutes.

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