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Chapter 10: Problem 34

Assuming that the two populations have unequal and unknown population standard deviations, construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) for the following. $$ \begin{array}{lll} n_{1}=48 & \bar{x}_{1}=.863 & s_{1}=.176 \\ n_{2}=46 & \bar{x}_{2}=.796 & s_{2}=.068 \end{array} $$

Short Answer

Expert verified
The 99% confidence interval for the difference of means is \( (0.009, 0.125). \)

Step by step solution

01

Compute the Difference of Sample Means

First calculate the difference of sample means: \(\bar{x}_{1}-\bar{x}_{2} = .863 - .796 = .067.\)
02

Compute the Margin of Error

Compute the margin of error using the formula for two samples margin of error: \(Z_{\alpha/2} * \sqrt{s_{1}^2/n_{1} + s_{2}^2/n_{2}}\). For 99% confidence level, the critical Z-score is approximately 2.576. So, the margin of error is \(2.576 * \sqrt{(.176)^2/48 + (.068)^2/46} = 0.058.\)
03

Calculate the Confidence Interval

Determine the confidence interval using the difference of sample means and the margin of error: \((\bar{x}_{1}-\bar{x}_{2}) 卤 \) Margin of Error = \(0.067 卤 0.058\). The 99% confidence interval for the difference of means is \((0.009, 0.125).\)

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Key Concepts

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

Sample Means
Sample means are the averages obtained from samples taken from a larger population. They are crucial in inferential statistics because they serve as estimates of the true population means. In practice, it鈥檚 rare to measure an entire population, so we rely on sample means to draw conclusions about the larger group.
  • The formula for a sample mean is \( \bar{x} = \frac{\sum x}{n} \), where \( \bar{x} \) is the sample mean, \( \sum x \) is the sum of all sample data points, and \( n \) is the number of data points (sample size).
  • In the exercise, \( \bar{x}_{1} = 0.863 \) and \( \bar{x}_{2} = 0.796 \). These are the means of the two samples being compared.
Sample means provide a foundation for calculating the confidence interval when we want to estimate the difference in means between two populations.
Margin of Error
The margin of error measures the extent of uncertainty around the sample mean in estimating the population mean. It indicates how much the sample results may deviate from the true population value.
  • The formula used to calculate the margin of error for two samples is \ Z_{\alpha/2} \sqrt{s_{1}^2/n_{1} + s_{2}^2/n_{2}} \, where \( Z_{\alpha/2} \) is the Z-score representing the confidence level, \( s \) is the sample standard deviation, and \( n \) is the sample size.
  • In our given solution, the critical Z-score for a 99% confidence level is 2.576. Applying this, we find the margin of error for the sample means to be approximately 0.058.
Understanding the margin of error helps us quantify the degree of confidence or reliability we have in the sample mean as an estimate of the population mean.
Z-score
Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It is expressed as a number of standard deviations from the mean. In confidence interval calculations, the Z-score translates the confidence level into a threshold that indicates how extreme our sample results are considered.
  • For a 99% confidence interval, the Z-score is approximately 2.576. This value is determined from the standard normal distribution and tells us that if we were to take many samples and construct confidence intervals, 99% of them would contain the true population mean.
  • The selection of the Z-score depends on the desired confidence level, with common options being 1.96 for 95%, 2.576 for 99%, and so on.
The Z-score is vital in helping us determine the margin of error, hence affecting the width and accuracy of our confidence intervals.
Population Standard Deviation
Population standard deviation is a measure that quantifies the amount of variation or dispersion in a population dataset. It reflects how data points spread around the mean, and it is important in calculating inferential statistics like confidence intervals.
  • In many real-world scenarios, the population standard deviation is unknown, which is the case in the example problem. We use sample standard deviations \( s_{1} \) and \( s_{2} \) as substitutes for the calculations.
  • Sample standard deviations help estimate the variability within the sample and provide an approximation for population dispersion, which is crucial for computing the margin of error.
Correctly estimating the population standard deviation or using sample values helps in accurate calculation of the confidence intervals, enhancing the reliability of our statistical inferences.

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Most popular questions from this chapter

Construct a \(95 \%\) confidence interval for \(p_{1}-p_{2}\) for the following. $$ n_{1}=100, \quad \hat{p}_{1}=.81, \quad n_{2}=150, \quad \hat{p}_{2}=.77 $$

What is the shape of the sampling distribution of \(\hat{p}_{1}-\hat{p}_{2}\) for two large samples? What are the mean and standard deviation of this sampling distribution?

According to Exercise \(10.27\), an insurance company wants to know if the average speed at which men drive cars is higher than that of women drivers. The company took a random sample of 27 cars driven by men on a highway and found the mean speed to be 72 miles per hour with a standard deviation of \(2.2\) miles per hour. Another sample of 18 cars driven by women on the same highway gave a mean speed of 68 miles per hour with a standard deviation of \(2.5\) miles per hour. Assume that the speeds at which all men and all women drive cars on this highway are both normally distributed with unequal population standard deviations. a. Construct a \(98 \%\) confidence interval for the difference between the mean speeds of cars driven by all men and all women on this highway. b. Test at a \(1 \%\) significance level whether the mean speed of cars driven by all men drivers on this highway is higher than that of cars driven by all women drivers. c. Suppose that the sample standard deviations were \(1.9\) and \(3.4\) miles per hour, respectively. Redo parts a and b. Discuss any changes in the results.

According to an estimate, the average earnings of female workers who are not union members are \(\$ 909\) per week and those of female workers who are union members are \(\$ 1035\) per week. Suppose that these average carnings are calculated based on random samples of 1500 female workers who are not union members and 2000 female workers who are union members. Further assume that the standard deviations for the two corresponding populations are \(\$ 70\) and \(\$ 90\), respectively. a. Construct a \(95 \%\) confidence interval for the difference between the two population means. b. Test at a \(2.5\) \% significance level whether the mean weekly earnings of female workers who are not union members are less than those of female workers who are union members.

A company that has many department stores in the southern states wanted to find at two such stores the percentage of sales for which at least one of the items was returned. A sample of 800 sales randomly selected from Store A showed that for 280 of them at least one item was returned. Another sample of 900 sales randomly selected from Store B showed that for 279 of them at least one item was returned. A. Construct a \(98 \%\) confidence interval for the difference between the proportions of all sales at the two stores for which at least one item is returned. b. Using a \(1 \%\) significance level, can you conclude that the proportions of all sales for which at least one item is returned is higher for Store A than for Store B?
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