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According to an estimate, the average earnings of female workers who are not union members are \(\$ 388\) per week and those of female workers who are union members are \(\$ 505\) per week. Suppose that these average earnings 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 \(\$ 30\) and \(\$ 35\), respectively. a. Construct a \(95 \%\) confidence interval for the difference between the two population means. b. Test at the \(2.5 \%\) 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.

Short Answer

Expert verified
The 95% confidence interval for the difference in mean earnings between the two groups of women and the result of the hypothesis testing will be found from the calculations in Step 4.

Step by step solution

01

Compute Confidence Interval

Use the formula for the confidence interval for the difference between population means: \[CI = (\overline{x_1} - \overline{x_2}) \pm Z * \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\], where \(\overline{x_1}\) and \(\overline{x_2}\) are the sample means, \(s_1\) and \(s_2\) are the sample standard deviations, and \(n_1\) and \(n_2\) are the sample sizes. Use the given values and Z-value for 95% confidence level, which is 1.96. This gives \[CI = (388 - 505) \pm 1.96 * \sqrt{\frac{30^2}{1500} + \frac{35^2}{2000}}\].
02

Compute Test Statistic

For the hypothesis testing, calculate the test statistic using the formula: \[Z = \frac{(\overline{x_1} - \overline{x_2}) - ( \mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]. This test is one-tailed, as the alternative hypothesis is testing if the mean earnings of non-union women is less than union women. So, \(\mu_1 - \mu_2\) = 0. Therefore, \[Z = \frac{(388 - 505) - 0}{\sqrt{\frac{30^2}{1500} + \frac{35^2}{2000}}}\].
03

Make Conclusion

Compare the computed test statistic with the critical value for 2.5% significance level in a one-tail test, which is -1.96. If the test statistic is less than the critical value, the null hypothesis is rejected.
04

Perform Calculations

Perform the calculations in the first two steps to get the numerical results for the confidence interval and test statistic.

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Key Concepts

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

Confidence Interval
A confidence interval provides a range of values that is believed to contain the true difference between two population means with a certain level of confidence, such as 95%. It is a critical concept in inferential statistics, allowing us to make predictions or inferences about a population based on sample data.
When constructing a confidence interval for the difference between two means, you'll often use the formula:\[ CI = (\overline{x_1} - \overline{x_2}) \pm Z * \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]Here:- \(\overline{x_1}\) and \(\overline{x_2}\) are the sample means.- \(s_1\) and \(s_2\) are sample standard deviations.- \(n_1\) and \(n_2\) represent sample sizes.- \(Z\) is the Z-value corresponding to the desired confidence level, like 1.96 for 95% confidence.
For instance, in our scenario, we calculate the confidence interval to estimate the true difference between the earnings of female workers, union versus non-union. If the interval does not include zero, it suggests a significant difference in means exists.
Hypothesis Testing
Hypothesis testing is used to determine whether there is enough statistical evidence in favor of a certain belief about a population parameter. This method can help you decide if a hypothesis about a population is likely to be true.
In the context of the exercise, we test if the mean earnings of non-union female workers are lower than those of union workers. The typical steps in hypothesis testing are:- **Formulate Hypotheses:** Define the null hypothesis (often that there is no effect or difference) and the alternative hypothesis (what you seek to prove).- **Calculate the Test Statistic:** Use the formula \[ Z = \frac{(\overline{x_1} - \overline{x_2}) - ( \mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]where \(\mu_1 - \mu_2\) is the hypothesized difference in the population means, often zero under null hypothesis.- **Determine Critical Value and Decision Rule:** Depending on the significance level (here, 2.5%), compare the test statistic to the critical value (-1.96 for a one-tailed test) to decide on the null hypothesis.- **Make a Conclusion:** Reject or fail to reject the null hypothesis based on the comparison.
For our problem, rejecting the null hypothesis would suggest that non-union female workers earn less on average than union workers.
Population Means
Understanding population means is fundamental in statistics, especially when comparing groups like in our exercise. The population mean is an average of a set of values for the entire population. However, since it's often impractical to gather data from an entire population, we use sample means to estimate the population mean.
In the context of our problem here, we are comparing two population means:- The mean weekly earnings of female workers who are not union members.- The mean weekly earnings of female workers who are union members.The exercise utilizes samples to estimate these population means since it's impractical to survey every member of each group.
Importantly, the sample means and standard deviations (\(\overline{x_1}\), \(\overline{x_2}\), \(s_1\), \(s_2\)) allow us to infer about the population means using formulas for both confidence intervals and hypothesis testing. With larger sample sizes (as in our example), estimates of population means become more accurate, providing better insights into the true averages.

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Most popular questions from this chapter

According to a June 2009 report (http://www.alertnet.org/thenews/newsdesk/L31011082.htm), \(68 \%\) of people with "green" jobs in North America felt that they had job security, whereas \(60 \%\) of people with green jobs in the United Kingdom felt that they had job security. Suppose that these results were based on samples of 305 people with green jobs from North America and 280 people with green jobs from the United Kingdom. a. Construct a \(96 \%\) confidence interval for the difference between the two population proportions. b. Using the \(2 \%\) significance level, can you conclude that the proportion of all people with green jobs in North America who feel that they have job security is higher than the corresponding proportion for the United Kingdom? Use the critical-value approach. c. Repeat part b using the \(p\) -value approach.

According to the information given in Exercise \(10.25\), a sample of 45 customers who drive luxury cars showed that their average distance driven between oil changes was 3187 miles with a sample standard deviation of \(42.40\) miles. Another sample of 40 customers who drive compact lower-price cars resulted in an average distance of 3214 miles with a standard deviation of \(50.70\) miles. Suppose that the standard deviations for the two populations are not equal. a. Construct a \(95 \%\) confidence interval for the difference in the mean distance between oil changes for all luxury cars and all compact lower-price cars. b. Using the \(1 \%\) significance level, can you conclude that the mean distance between oil changes is lower for all luxury cars than for all compact lower-price cars? c. Suppose that the sample standard deviations were \(28.9\) and \(61.4\) miles, respectively. Redo parts a and b. Discuss any changes in the results.

A car magazine is comparing the total repair costs incurred during the first three years on two sports cars, the T-999 and the XPY. Random samples of 45 T-999s and 51 XPYs are taken. All 96 cars are 3 years old and have similar mileages. The mean of repair costs for the 45 T-999 cars is \(\$ 3300\) for the first 3 years. For the 51 XPY cars, this mean is \(\$ 3850 .\) Assume that the standard deviations for the two populations are \(\$ 800\) and \(\$ 1000\), respectively. a. Construct a \(99 \%\) confidence interval for the difference between the two population means. b. Using the \(1 \%\) significance level, can you conclude that such mean repair costs are different for these two types of cars? c. What would your decision be in part \(\mathrm{b}\) if the probability of making a Type I error were zero? Explain.

A June 2009 Gallup Poll asked a sample of Americans whether they trust specific groups or individuals when it comes to making recommendations about healthcare reform. Sixty percent of Democrats and \(68 \%\) of Republicans stated that they trust doctors' opinions about healthcare reform (Source: http://www.gallup.com/poll/120890/Healthcare-Americans-Trust-Physicians- Politicians.aspx). Suppose this survey included 340 Democrats and 306 Republicans. a. Make a \(90 \%\) confidence interval for the difference in the population proportions for the two groups of people. b. At the \(5 \%\) significance level, can you conclude that the proportion of all Democrats who trust doctors' opinions about healthcare reform differs from the proportion of all Republicans who trust doctors' opinions about healthcare reform?

A sample of 500 observations taken from the first population gave \(x_{1}=305\). Another sample of 600 observations taken from the second population gave \(x_{2}=348\). a. Find the point estimate of \(p_{1}-p_{2}\). b. Make a \(97 \%\) confidence interval for \(p_{1}-p_{2}\). c. Show the rejection and nonrejection regions on the sampling distribution of \(\hat{p}_{1}-\hat{p}_{2}\) for \(H_{0}: p_{1}=p_{2}\) versus \(H_{1}: p_{1}>p_{2} .\) Use a significance level of \(2.5 \% .\) d. Find the value of the test statistic \(z\) for the test of part \(\mathrm{c}\). e. Will you reject the null hypothesis mentioned in part \(\mathrm{c}\) at a significance level of \(2.5 \%\) ?

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