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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.

Short Answer

Expert verified
The solution involves calculating the standard error, constructing a 95% confidence interval for the difference in means, performing a hypothesis test, and interpreting the result. The exact values of the confidence interval and the decision of the hypothesis test will depend on the calculations.

Step by step solution

01

Calculate the Standard Error

The standard error (SE) is calculated using the formula SE = sqrt((SD1^2/n1) + (SD2^2/n2)), where SD1 and SD2 are the standard deviations of the two populations, and n1 and n2 are the sample sizes. Here, SD1 = \$70, n1 = 1500 (for female workers who are not union members), SD2 = \$90 and n2 = 2000 (for female workers who are union members). So the standard error would be: SE = sqrt((70^2/1500) + (90^2/2000)).
02

Construct the 95% Confidence Interval

The formula for the confidence interval is (X1 - X2) ± (Z * SE). Here X1 and X2 are the sample means, and Z is the Z-value from the standard normal distribution for the desired confidence level. For a 95% confidence level, the Z-value is approximately 1.96. Here, X1 = \$909 and X2 = \$1035. The 95% confidence interval for the difference in population means is therefore (\$909 - \$1035) ± 1.96 * SE.
03

Conduct a Hypothesis Test

The hypotheses for this test are: \nNull hypothesis (H0): µ1 - µ2 ≥ 0 (the mean weekly earnings of female workers who are not union members is equal to or greater than that of union workers) \nAlternative hypothesis (H1): µ1 - µ2 < 0 (the mean weekly earnings of female workers who are not union members is less than that of union workers)\nThe test statistic for this test is given by (X1 - X2) / SE = (\$909 - \$1035) / SE\nWe compare the test statistic to the critical value of Z for a one-tailed test at a 2.5% significance level. The critical value of Z for a left-tailed test at a 2.5% significance level is approximately -1.96.
04

Make a Decision and Interpret the Result

If the calculated test statistic is less than the critical value, we reject the null hypothesis. Otherwise, we do not reject the null hypothesis. The interpretation of the result will be along the lines of ‘there is sufficient evidence at the 2.5% significance level to support the claim that the mean weekly earnings of female workers who are not union members are less than those of female workers who are union members’, or 'the evidence does not support the claim'.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a way to determine whether there is enough evidence in a sample of data to support a specific claim about a population parameter. In our example, we test if the mean earnings of non-union female workers are less than those of union female workers. - **Null Hypothesis (H0):** This is a statement of no effect or no difference. For our problem, it states that the mean earnings of non-union female workers are equal to or greater than that of union workers. - **Alternative Hypothesis (H1):** Proposes that there is an effect or a difference. Here, it claims that non-union female workers earn less. During hypothesis testing, we calculate a test statistic to help determine if the null hypothesis should be rejected. We also decide on significance level (like 2.5%) which helps in comparing the test statistic against a critical value. If the test statistic falls into the critical region (beyond a threshold), we have significant evidence to reject the null hypothesis. This process shows whether our sample data provides enough statistical evidence to support the alternative hypothesis.
Population Mean Difference
The population mean difference refers to the expected difference between two population means. In our context, it's about the weekly earnings difference between female workers who are union members versus those who are not. Estimating this difference is critical in understanding if one group indeed earns more than the other.- **Sample Means (X1 and X2):** These are calculated averages from our samples. Here, non-union workers average \(\\(909\) per week, and union workers average \(\\)1035\). The essence of analyzing the population mean difference is to clarify how wide or narrow that difference might be in practical terms. A substantial mean difference, statistically validated, could imply genuine disparities in earnings that stem from being union or non-union. Furthermore, confidence intervals are used to provide a range for the potential difference, offering more insight than single-point estimates.
Standard Error
The standard error (SE) measures how much your sample estimate of the population mean is expected to vary due to random sampling. It's crucial for constructing confidence intervals and testing hypotheses about population differences. - **Standard Error Formula:** \[ SE = \sqrt{\left(\frac{SD1^2}{n1}\right) + \left(\frac{SD2^2}{n2}\right)} \] Where: - \(SD1\) and \(SD2\) are the population standard deviations. - \(n1\) and \(n2\) are the sample sizes.In our example, it helps determine how much the difference in weekly earnings might vary due to sampling variability. A smaller SE suggests more reliable mean estimates, increasing confidence in our hypothesis test results. It is pivotal for precise interval calculations such as the confidence interval, outlining the expected range of the population mean difference.
Significance Level
The significance level ( α ) is a threshold set by the researcher before performing hypothesis testing. It determines the probability of rejecting the null hypothesis when it is actually true (Type I error). In our problem, the level is set at 2.5%. - **Choosing a Significance Level:** - A common choice is 5% , but more stringent levels (like 2.5% ) are used when we want less risk of false positives. - **Critical Value and Decision:** - For a one-tailed test like ours, where we check if one mean is less than another, the critical Z value is -1.96. - If the test statistic is less than this critical value, we reject the null hypothesis. The significance level plays a pivotal role in ensuring that the conclusions drawn from data hold robust statistical meaning. Lower significance levels opportunely lower the Type I error risk, enhancing trust in the affirmative results of alternative hypotheses.

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Most popular questions from this chapter

The following information was obtained from two independent samples selected from two normally distributed populations with unknown but equal standard deviations. \(\begin{array}{llllllllllllll}\text { Sample 1: } & 47.7 & 46.9 & 51.9 & 34.1 & 65.8 & 61.5 & 50.2 & 40.8 & 53.1 & 46.1 & 47.9 & 45.7 & 49.0 \\\ \text { Sample 2: } & 50.0 & 47.4 & 32.7 & 48.8 & 54.0 & 46.3 & 42.5 & 40.8 & 39.0 & 68.2 & 48.5 & 41.8 & \end{array}\) a. Let \(\mu_{1}\) be the mean of population 1 and \(\mu_{2}\) be the mean of population \(2 .\) What is the point estimate of \(\mu_{1}-\mu_{2} ?\) b. Construct a \(98 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) c. Test at a \(1 \%\) significance level if \(\mu_{1}\) is greater than \(\mu_{2}\).

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 a \(1 \%\) 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.

According to a report in The New York Times, in the United States, accountants and auditors earn an average of \(\$ 70,130\) a year and loan officers carn \(\$ 67,960\) a year (Jessica Silver-Greenberg, The New York Times, April 22,2012 ). Suppose that these estimates are based on random samples of 1650 accountants and auditors and 1820 loan officers. Further assume that the sample standard deviations of the salaries of the two groups are \(\$ 14,400\) and \(\$ 13,600\), respectively, and the population standard deviations are equal for the two groups. a. Construct a \(98 \%\) confidence interval for the difference in the mean salaries of the two groupsaccountants and auditors, and loan officers. b. Using a \(1 \%\) significance level, can you conclude that the average salary of accountants and auditors is higher than that of loan officers?

A state that requires periodic emission tests of cars operates two emission test stations, \(\mathrm{A}\) and \(\mathrm{B}\), in one of its towns. Car owners have complained of lack of uniformity of procedures at the two stations, resulting in different failure rates. A sample of 400 cars at Station A showed that 53 of those failed the test; a sample of 470 cars at Station \(\mathrm{B}\) found that 51 of those failed the test. a. What is the point estimate of the difference between the two population proportions? b. Construct a 95\% confidence interval for the difference between the two population proportions. c. Testing at a \(5 \%\) significance level, can you conclude that the two population proportions are different? Use both the critical-value and the \(p\) -value approaches.

The following information is obtained from two independent samples selected from two normally distributed populations. $$ \begin{array}{lll} n_{1}=18 & \bar{x}_{1}=7.82 & \sigma_{1}=2.35 \\ n_{2}=15 & \bar{x}_{2}=5.99 & \sigma_{2}=3.17 \end{array} $$ a. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). Find the margin of error for this estimate.

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