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A sociologist studying New York City ethnic groups wants to determine if there is a difference in income for immigrants from four different countries during their first year in the city. She obtained the data in the following table from a random sample of immigrants from these countries (incomes in thousands of dollars). Use a \(0.05\) level of significance to test the claim that there is no difference in the earnings of immigrants from the four different countries. $$ \begin{array}{rrcr} \text { Country I } & \text { Country II } & \text { Country III } & \text { Country IV } \\ 12.7 & 8.3 & 20.3 & 17.2 \\ 9.2 & 17.2 & 16.6 & 8.8 \\ 10.9 & 19.1 & 22.7 & 14.7 \\ 8.9 & 10.3 & 25.2 & 21.3 \\ 16.4 & & 19.9 & 19.8 \end{array} $$

Short Answer

Expert verified
After conducting an ANOVA test, compare the F-statistic to the critical value to determine if there is a significant income difference between the groups.

Step by step solution

01

Set Up Hypotheses

We need to set up the null and alternative hypotheses for the analysis. The null hypothesis () claims there is no difference in average incomes among the groups (\(H_0: \mu_1 = \mu_2 = \mu_3 = \mu_4\)). The alternative hypothesis (\(H_a\)) states at least one group mean is different (\(H_a: \text{At least one } \mu \text{ is different}\)).
02

Choose the Significance Level and Test

We are using a significance level of \(\alpha = 0.05\) for this test. The appropriate test for comparing more than two means is the ANOVA (Analysis of Variance) test, which will help us determine if there is a significant difference between the group means.
03

Organize the Data

We have the income data from four different countries. Begin by organizing this data: - Country I: 12.7, 9.2, 10.9, 8.9, 16.4 - Country II: 8.3, 17.2, 19.1, 10.3 - Country III: 20.3, 16.6, 22.7, 25.2, 19.9 - Country IV: 17.2, 8.8, 14.7, 21.3, 19.8
04

Calculate Group Means and Totals

Calculate the mean for each group:\[\begin{align*}\text{Country I Mean} &= \frac{12.7 + 9.2 + 10.9 + 8.9 + 16.4}{5} = 11.62 \\text{Country II Mean} &= \frac{8.3 + 17.2 + 19.1 + 10.3}{4} = 13.725 \\text{Country III Mean} &= \frac{20.3 + 16.6 + 22.7 + 25.2 + 19.9}{5} = 20.94 \\text{Country IV Mean} &= \frac{17.2 + 8.8 + 14.7 + 21.3 + 19.8}{5} = 16.36 \\end{align*}\]
05

Conduct ANOVA Test

Using the means calculated, conduct the ANOVA test. Calculate the "Between Group" and "Within Group" variations to find the F-statistic. Use \\(F = \frac{\text{Mean Square Between (MSB)}}{\text{Mean Square Within (MSW)}}\) to find the F-statistic value. This requires summing squared deviations for each group and the overall mean.
06

Determine the ANOVA F Critical Value

Consult the F distribution table to find the critical value for \(F\) at \(\alpha = 0.05\) with degrees of freedom (df1, df2) corresponding to the number of groups and total data points minus the number of groups. Typically, with 3 and 15 df's, find \(F_{critical}\).
07

Make a Conclusion

Compare the calculated F-statistic with the critical F value:- If \(F_{calculated} > F_{critical}\), reject \(H_0\) (at least one mean is different).- If \(F_{calculated} \leq F_{critical}\), do not reject \(H_0\) (no significant difference).

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Key Concepts

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

Understanding Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions or inferences about population parameters based on sample data. In the context of comparing the income of immigrants from four different countries, we start by establishing two hypotheses.
  • Null Hypothesis (\(H_0\)): This hypothesis assumes no difference in average incomes among the groups. Mathematically, we express it as \(H_0: \mu_1 = \mu_2 = \mu_3 = \mu_4\). This statement suggests that all groups have the same mean income.
  • Alternative Hypothesis (\(H_a\)): This hypothesis proposes that at least one group's mean is different. It's formulated as \(H_a:\) At least one \(\mu\) is different. This implies variation among the group's incomes.
To test these hypotheses, we use an ANOVA test, which can assess differences across more than two groups. Each hypothesis sets the stage for the subsequent analysis, and the outcome will hinge on whether the data provides enough evidence to reject the null hypothesis.
The Role of Significance Level in Testing
The significance level, often denoted as \(\alpha\), is a critical threshold in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true (a Type I error). Setting this threshold helps control the risk of making this kind of mistake.In the case of our sociologist's study, an \(\alpha = 0.05\) level is chosen. Here's why this level is important:
  • Risk Management: A 0.05 significance level means there is a 5% risk of incorrectly concluding that there is a difference in incomes when there is none.
  • Standard Practice: The 5% threshold is widely used in scientific research as a standard level for decision-making. It balances the risk of Type I errors against the need for sensitivity in detecting true differences.
  • Interpreting Results: The outcome of the ANOVA test is compared to this critical value: if the calculated statistic exceeds the 0.05 threshold, we have grounds to reject the null hypothesis and claim a significant difference.
This significance level serves as a line in the sand – a benchmark that guides the researcher in making an informed decision about the hypotheses.
Diving Into Mean Comparison
Mean comparison is a crucial part of analyzing group differences, especially when deploying an ANOVA test. For our exercise, this involves calculating the average income for immigrants from each country and comparing these means to determine disparities.Let's explore how it's applied:
  • Calculating Means: First, we compute the mean income for each country's data. This sets the groundwork for comparing across groups. For example, Country I has a mean income of 11.62, and Country III, 20.94.
  • Between and Within Groups Variation: ANOVA analyzes differences by separating them into two parts—between-group and within-group variations. The between-group variation measures disparities among the different country's means. In contrast, the within-group variation looks at differences within each country's data.
  • Using ANOVA: By applying the formula \(F = \frac{\text{Mean Square Between (MSB)}}{\text{Mean Square Within (MSW)}}\), we determine if the observed differences in means are statistically significant by comparing the calculated F-statistic against a critical value from the F-distribution table.
Through this detailed comparison, the ANOVA test provides a robust way to decide if any income differences are statistically significant, thereby facilitating informed conclusions.

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Most popular questions from this chapter

The quantity of dissolved oxygen is a measure of water pollution in lakes, rivers, and streams. Water samples were taken at four different locations in a river in an effort to determine if water pollution varied from location to location. Location I was 500 meters above an industrial plant water discharge point and near the shore. Location II was 200 meters above the discharge point and in midstream. Location III was 50 meters downstream from the discharge point and near the shore. Location IV was 200 meters downstream from the discharge point and in midstream. The following table shows the results. Lower dissolved oxygen readings mean more pollution. Because of the difficulty in getting midstream samples, ecology students collecting the data had fewer of these samples. Use an \(\alpha=0.05\) level of significance. Do we reject or not reject the claim that the quantity of dissolved oxygen does not vary from one location to another? $$ \begin{array}{cccc} \text { Location I } & \text { Location II } & \text { Location III } & \text { Location IV } \\ 7.3 & 6.6 & 4.2 & 4.4 \\ 6.9 & 7.1 & 5.9 & 5.1 \\ 7.5 & 7.7 & 4.9 & 6.2 \\ 6.8 & 8.0 & 5.1 & \\ 6.2 & & 4.5 & \end{array} $$

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