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Chapter 9: Problem 9

Driving (Example 1) Drivers in Wyoming drive more miles yearly than motorists in any other state. The annual number of miles driven per licensed driver in Wyoming is 22,306 miles. Assume the standard deviation is 5500 miles. A random sample of 200 licensed drivers in Wyoming is selected and the mean number of miles driven yearly for the sample is calculated. (Source: 2017 World Almanac and Book of Facts) a. What value would we expect for the sample mean? b. What is the standard error for the sample mean?

Short Answer

Expert verified
The expected value for the sample mean is 22306 miles and the standard error for the sample mean is approximately 389 miles.

Step by step solution

01

Calculate Expected Value for Sample Mean

The expected value for the sample mean is equal to the population mean. So, the expected value for the sample mean is 22306 miles.
02

Calculate the Standard Error for the Sample Mean

The standard error for the sample mean is calculated by dividing the population standard deviation by the square root of the sample size. So, the standard error for the sample mean is 5500 miles divided by the square root of 200, which is \( \frac{5500}{\sqrt{200}} \) miles.
03

Calculate the value

Now using a calculator or computation software, compute the value of \( \frac{5500}{\sqrt{200}} \).

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Key Concepts

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

Sample Mean
The sample mean is a statistical measure that provides us with an average by summing all data points in a sample and dividing by the number of observations. It's a way to express the central value in a sample and is used to draw inferences about the population from which the sample is drawn. In the context of the Wyoming drivers example:
  • The sample mean gives us an idea of how many miles, on average, the drivers in the sample have driven.
  • We expect the sample mean to be close to the population mean, particularly with large samples. This provides validity to using the sample as an estimate for the larger group.
Understanding the sample mean helps us gauge the data's consistency and ensure our sample accurately reflects the population's driving patterns.
Standard Deviation
Standard deviation is a measure of variability or dispersion within a dataset. It represents how much individual data points typically deviate from the mean. In the driving data scenario:
  • The given standard deviation is 5500 miles, indicating the extent of variation from the average miles driven.
  • Higher standard deviation means more variability in the number of miles driven among drivers.
  • It provides critical insight, especially when comparing different groups, as it tells us how spread out the drivers' annual driving distances are.
Knowing the standard deviation helps us understand the reliability of the mean and the variation we might expect in additional samples.
Standard Error
The standard error is a vital concept in statistics, providing an estimate of the variability of the sample mean if repeated samples were taken. It's calculated using the formula:\[ \text{Standard Error} = \frac{\text{Population Standard Deviation}}{\sqrt{\text{Sample Size}}} \]For the Wyoming drivers' data:
  • The population standard deviation is 5500 miles.
  • The sample size is 200 drivers.
Therefore, the standard error is computed as \( \frac{5500}{\sqrt{200}} \). This value, the standard error, shrinks as the sample size increases, signifying more precision in estimating the population mean from the sample mean.
Population Mean
The population mean is a measure of central tendency that describes the average of all data points in a complete population. It's a key parameter in statistics:
  • For the Wyoming drivers, the population mean is 22,306 miles. This is the actual average number of miles driven per driver in the entire state, often considered a true representation of the population.
  • In practice, the population mean is rarely known, hence the reliance on sample means and other statistical methods to estimate it.
  • It's foundational for comparing samples against the larger population, helping us judge the accuracy of sample-based estimates.
Understanding the population mean aids in making informed predictions and decisions based on representative data, especially in large datasets.

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Most popular questions from this chapter

Potatoes The weights of four randomly and independently selected bags of potatoes labeled 20 pounds were found to be \(21.0\), \(21.5,20.5\), and \(21.2\) pounds. Assume Normality. a. Find a \(95 \%\) confidence interval for the mean weight of all bags of potatoes. b. Does the interval capture \(20.0\) pounds? Is there enough evidence to reject a mean weight of 20 pounds?

Why Is \(n-1\) in the Sample Standard Deviation? Why do we calculate \(s\) by dividing by \(n-1\), rather than just \(n\) ? $$ s^{2}=\frac{\sum(x-\bar{x})^{2}}{n-1} $$ The reason is that if we divide by \(n-1\), then \(s^{2}\) is an unbiased estimator of \(\sigma^{2}\), the population variance. We want to show that \(s^{2}\) is an unbiased estimator of \(\sigma^{2}\), sigma squared. The mathematical proof that this is true is beyond the scope of an introductory statistics course, but we can use an example to demonstrate that it is. First we will use a very small population that consists only of these three numbers: 1,2, and 5 . You can determine that the population standard deviation, \(\sigma\), for this population is \(1.699673\) (or about \(1.70\) ), as shown in the \(\mathrm{TI}-84\) output. So the population variance, sigma squared, \(\sigma^{2}\), is \(2.888889\) (or about 2.89). Now take all possible samples, with replacement, of size 2 from the population, and find the sample variance, \(s^{2}\), for each sample. This process is started for you in the table. Average these sample variances \(\left(s^{2}\right)\), and you should get approximately \(2.88889 .\) If you do. then you have demonstrated that \(s^{2}\) is an unbiased estimator of \(\sigma^{2}\), sigma squared. Show your work by filling in the accompanying table and show the average of \(s^{2}\).

Men's Pulse Rates (Example 10) A random sample of 25 men's resting pulse rates shows a mean of 72 beats per minute and \(\mathrm{a}\) standard deviation of 13 . a. Find a \(95 \%\) confidence interval for the population mean pulse rate for men, and report it in a sentence. You may use the table given for Exercise \(9.25\) b. Find a \(99 \%\) confidence interval. c. Which interval is wider, and why?

Pulse Difference The following numbers are the differences in pulse rate (beats per minute) before and after running for 12 randomly selected people. $$ 24,12,14,12,16,10,0,4,13,42,4, \text { and } 16 $$ Positive numbers mean the pulse rate went up. Test the hypothesis that the mean difference in pulse rate was more than 0 , using a significance level of \(0.05 .\) Assume the population distribution is Normal.

Atkins Diet Difference Ten people went on an Atkins diet for a month. The weight losses experienced (in pounds) were $$ 3,8,10,0,4,6,6,4,2, \text { and }-2 $$ The negative weight loss is a weight gain. Test the hypothesis that the mean weight loss was more than 0 , using a significance level of \(0.05\). Assume the population distribution is Normal.
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