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According to an estimate, the average earnings of female workers who are not union members are \(\$ 1120\) per week and those of female workers who are union members are \(\$ 1244\) 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 \(\$ 70\) 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 is less than that of female workers who are union members.

Short Answer

Expert verified
The 95% confidence interval for the difference between the average weekly earnings of the two groups of female workers is \[-130.38, -117.62\]. Indeed, it has been statistically proved with a significance level of 2.5% that the average weekly earning of female workers who are not union members is less than that of female workers who are union members.

Step by step solution

01

Calculate mean difference

Calculate the mean difference between the two groups, it is \(don = 1120 - 1244 = -124\). Here, don represents the mean difference.
02

Calculate standard error

Compute the standard error (SE). The formula for the standard error is\[SE = \sqrt{(\frac{s_1 ^2}{n_1}+\frac{s_2 ^2}{n_2})}\] Where \(s_1\) and \(s_2\) are standard deviations and \(n_1\) and \(n_2\) are the sizes of the two groups. By plugging in the values, we get\[SE = \sqrt{(\frac{70^2}{1500}+\frac{90^2}{2000})} = 3.2533...\]
03

Construct 95% confidence interval

A 95% confidence interval is \(\bar{X} \pm 1.96 * SE\). By substituting the values we calculated before we get \[-124 \pm 1.96 * 3.25\], which results in a 95% confidence interval of \[-130.38, -117.62\]. This means we can be 95% confident that the the true difference of means is within this range.
04

Hypothesis testing

In this step, we need to test if the mean weekly earnings of female workers who are not union members are less than that of union members at a 2.5% significance level. This is a one-tailed test since we are checking for 'less than'. The null hypothesis is that the means are equal i.e. \(\bar{X}= 0\). The alternative hypothesis is that the mean of non-union members is less than that of union members, i.e. \(\bar{X} < 0\). Calculate the Z-score using the formula \[Z = \frac{\bar{X} - \mu}{SE}\], where \(\mu\) is the population mean which is 0 and \(\bar{X}\) and SE we calculated before. This gives us a Z value of \frac{-124 - 0}{3.25} = -38.1538. Looking this up in Z table gives us a p value close to 0. Since the p value is less than the 2.5% significance level, we reject the null hypothesis and conclude that the mean weekly earnings of female workers who are not union members is less than that of female workers who are union members.

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Key Concepts

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

Standard Error
The standard error (SE) is a key component in statistics, especially when dealing with sample data. It helps us understand the variability or spread of sample means around the true population mean. In simpler terms, the standard error gives us an idea of how much the sample mean is expected to fluctuate from one sample to another. This is particularly important when making inferences about the population from which the samples are drawn.

For any given dataset, the formula for standard error is:
  • \( SE = \sqrt{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)} \)
where \(s_1\) and \(s_2\) are the standard deviations of the samples, and \(n_1\) and \(n_2\) are their respective sizes. You can think of the standard error as a measure of precision; a smaller SE indicates a more precise estimate of the population mean difference. By understanding the SE, you're essentially getting insight into the reliability of your sample statistics as an estimate of the population parameter.
Hypothesis Testing
Hypothesis testing is a fundamental method used in statistics to make decisions or inferences about populations based on sample data. It involves comparing your findings to what would be expected under a certain hypothesis.
  • The Null Hypothesis (\( H_0 \)): This is usually a statement of no effect or no difference. In the given exercise, \( H_0 \) suggests there is no difference in mean earnings between the two groups of female workers.
  • The Alternative Hypothesis (\( H_a \)): This is what you want to establish. It indicates the presence of an effect or difference. Here, \( H_a \) suggests the mean earnings of non-union members are less than those of union members.
Once you set up these hypotheses, you calculate a test statistic, like the Z-score, which helps determine how far away your sample result is from what you expect if the null hypothesis were true. Using statistical tables or software, you then find a p-value. If this p-value is less than the chosen significance level (in this case, 2.5%), you reject the null hypothesis and accept that there's a statistically significant difference.
Mean Difference
The mean difference is a simple yet powerful statistic that represents the difference between the average values of two data groups. In practice, it allows us to measure the effect size or magnitude of the difference, which is essential for comparing two groups.

To find the mean difference, you subtract the mean of one group from the mean of the other. For example, in the exercise, the mean earnings of non-union female workers is \( \\(1120 \) and for union female workers it's \( \\)1244 \). Hence, the mean difference is:
  • \( 1120 - 1244 = -124 \)
This negative mean difference indicates that, on average, non-union workers earn \( \$124 \) less than union workers. It’s important to consider this number with other statistics like standard error and confidence intervals, as they give context to the reliability and significance of the mean difference observed.

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Most popular questions from this chapter

Quadro Corporation has two supermarket stores in a city. The company's quality control department wanted to check if the customers are equally satisfied with the service provided at these two stores. A sample of 380 customers selected from Supermarket I produced a mean satisfaction index of \(7.6\) (on a scale of 1 to 10,1 being the lowest and 10 being the highest) with a standard deviation of \(.75 .\) Another sample of 370 customers selected from Supermarket II produced \(a\) mean satisfaction index of \(8.1\) with a standard deviation of \(.59\). Assume that the customer satisfaction index for each supermarket has unknown but same population standard deviation. a. Construct a \(98 \%\) confidence interval for the difference between the mean satisfaction indexes for all customers for the two supermarkets. b. Test at a \(1 \%\) significance level whether the mean satisfaction indexes for all customers for the two supermarkets are different.

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Manufacturers of two competing automobile models, Gofer and Diplomat, each claim to have the lowest mean fuel consumption. Let \(\mu_{1}\) be the mean fuel consumption in miles per gallon (mpg) for the Gofer and \(\mu_{2}\) the mean fuel consumption in mpg for the Diplomat. The two manufacturers have agreed to a test in which several cars of each model will be driven on a 100 -mile test run. The fuel consumption, in \(\mathrm{mpg}\), will be calculated for each test run. The average mpg for all 100 -mile test runs for each model gives the corresponding mean. Assume that for each model the gas mileages for the test runs are normally distributed with \(\sigma=2 \mathrm{mpg}\). Note that each car is driven for one and only one 100 -mile test run. a. How many cars (i.e., sample size) for each model are required to estimate \(\mu_{1}-\mu_{2}\) with a \(90 \%\) confidence level and with a margin of error of estimate of \(1.5 \mathrm{mpg}\) ? Use the same number of cars (i.e., sample size) for each model. b. If \(\mu_{1}\) is actually \(33 \mathrm{mpg}\) and \(\mu_{2}\) is actually \(30 \mathrm{mpg}\), what is the probability that five cars for each model would yield \(\bar{x}_{1} \geq \bar{x}_{2}\) ?

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