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Chapter 8: Problem 12

Use the following information to answer the next seven exercises: The U.S. Census Bureau conducts a study to determine the time needed to complete the short form. The Bureau surveys 200 people. The sample mean is 8.2 minutes. There is a known standard deviation of 2.2 minutes. The population distribution is assumed to be normal. Suppose the Census needed to be 98% confident of the population mean length of time. Would the Census have to survey more people? Why or why not?

Short Answer

Expert verified
No, the Census does not need to survey more people as the current sample provides a reasonable confidence interval with a small margin of error at a 98% confidence level.

Step by step solution

01

Understanding Confidence Intervals

A confidence interval gives a range of values for a population parameter like the mean. Here, we want a 98% confidence interval for the mean time to complete the form.
02

Identify Values

The sample mean is \( \bar{x} = 8.2 \) minutes, and the known standard deviation \( \sigma = 2.2 \) minutes. The sample size \( n = 200 \). We will use 98% confidence level for our calculations.
03

Find the Z-Score for 98% Confidence Level

For a 98% confidence level, the Z-score (critical value) that corresponds to the middle 98% of a standard normal distribution is approximately 2.33.
04

Calculate the Margin of Error

The margin of error (ME) is calculated using the formula: \[ ME = Z \times \left(\frac{\sigma}{\sqrt{n}}\right) \]Substitute the values: \( ME = 2.33 \times \left(\frac{2.2}{\sqrt{200}}\right) \). Calculate the standard error first: \( \frac{2.2}{\sqrt{200}} \approx 0.1556 \). Then, \( ME = 2.33 \times 0.1556 \approx 0.3626 \).
05

Analyze the Confidence Interval

The 98% confidence interval for the population mean is given by: \[ \bar{x} \pm ME = 8.2 \pm 0.3626 \]This gives us the interval: \( (7.8374, 8.5626) \).
06

Determine if a Larger Sample is Needed

The standard deviation and confidence interval show that with 200 people surveyed, we achieve a reasonably small margin of error at this confidence level. Lowering the margin of error further would require a larger sample size. However, the current sample size provides a relatively narrow interval given a high confidence level.

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Key Concepts

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

Sample Mean
Imagine you've collected time data from 200 people on how long they take to complete a census form. The 'sample mean' is essentially the average time calculated from this data group. In this exercise, the sample mean is noted as 8.2 minutes.

This value serves as a point estimate of the population mean, which represents the average time for the entire population. However, without surveying every individual, there is uncertainty. That's where confidence intervals come in, providing a range in which the true population mean is likely to fall with a certain level of confidence, such as 98% in our case.
Standard Deviation
Standard deviation is a measure that quantifies the amount of variation or dispersion in a set of data values. When the standard deviation is small, it indicates that the data points are close to the mean, while a larger standard deviation indicates more spread out data points.

In our study, the standard deviation is 2.2 minutes. This indicates how much individual times deviate from the average of 8.2 minutes. It's an assumption that provides a basis for further calculations, helping to define how much our sample mean could vary from the actual population mean.
Margin of Error
The margin of error is an element of the confidence interval that describes the range within which the true population parameter is expected to fall. For example, in this context, it shows how much the average survey response might differ from the actual average completion time of the entire population.

To calculate the margin of error, you need the standard deviation, sample size, and confidence level. For our data, the formula for margin of error is:
  • Calculate the standard error by dividing the standard deviation (2.2) by the square root of the sample size (200): \( \frac{2.2}{\sqrt{200}} \approx 0.1556 \).
  • Multiply the result by the Z-score for the 98% confidence level, which is 2.33:
    \( ME = 2.33 \times 0.1556 \approx 0.3626 \).
Thus, the margin of error gives a buffer around the sample mean, creating the confidence interval \( (7.8374, 8.5626) \), indicating where the true population mean likely falls.

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Most popular questions from this chapter

Use the following information to answer the next five exercises. A hospital is trying to cut down on emergency room wait times. It is interested in the amount of time patients must wait before being called back to be examined. An investigation committee randomly surveyed 70 patients. The sample mean was 1.5 hours with a sample standard deviation of 0.5 hours. Identify the following: a. \(x=\) b. \(s_{x}=\) C. \(n=\) d. \(n-1=\)

Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 pounds. We wish to construct a 95% confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 pounds. The sample standard deviation is 11 pounds. Which distribution should you use for this problem?

Suppose that a committee is studying whether or not there is waste of time in our judicial system. It is interested in the mean amount of time individuals waste at the courthouse waiting to be called for jury duty. The committee randomly surveyed 81 people who recently served as jurors. The sample mean wait time was eight hours with a sample standard deviation of four hours. a. i. \(\overline{x}=\) _____ ii. \(s_{x}=\) _____ iii. \(n=\) _____ iv. \(n-1=\) _____ b. Define the random variables \(X\) and \(\overline{X}\) in words. c. Which distribution should you use for this problem? Explain your choice. d. Construct a 95\(\%\) confidence interval for the population mean time wasted. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. Explain in a complete sentence what the confidence interval means.

A camp director is interested in the mean number of letters each child sends during his or her camp session. The population standard deviation is known to be 2.5. A survey of 20 campers is taken. The mean from the sample is 7.9 with a sample standard deviation of 2.8. a. i. \(\overline{x}=\) _____ ii. \(\sigma=\) _____ iii. \(n=\) _____ b. Define the random variables \(X\) and \(\overline{X}\) in words. c. Which distribution should you use for this problem? Explain your choice. d. Construct a 90\(\%\) confidence interval for the population mean number of letters campers send home. i. State the confidence interval. ii. Sketch the graph. iil. Calculate the error bound. e. What will happen to the error bound and confidence interval if 500 campers are surveyed? Why?

Use the following information to answer the next ten exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 pounds with a standard deviation of 0.1 pounds. The population standard deviation is known to be 0.2 pounds. Construct a 90% confidence interval for the population mean weight of the heads of lettuce. State the confidence interval, sketch the graph, and calculate the error bound.
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