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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
Perform an ANOVA; if F-statistic > critical value, reject the null hypothesis. Otherwise, do not reject it.

Step by step solution

01

Define the Hypotheses

The null hypothesis, \( H_0 \), states that there is no difference in the mean incomes of immigrants from the four countries. The alternative hypothesis, \( H_1 \), states that at least one country's mean income is different. We will use an ANOVA test to compare the means.
02

Set the Level of Significance

The level of significance, \( \alpha \), is 0.05. This means there is a 5% risk of concluding that there is a difference in mean incomes when there actually is none.
03

Calculate Group Means

Compute the mean income for each country. For Country I, the mean is \( \frac{12.7 + 9.2 + 10.9 + 8.9 + 16.4}{5} = 11.62 \). For Country II, the mean is \( \frac{8.3 + 17.2 + 19.1 + 10.3}{4} = 13.725 \). For Country III, the mean is \( \frac{20.3 + 16.6 + 22.7 + 25.2 + 19.9}{5} = 20.94 \). For Country IV, the mean is \( \frac{17.2 + 8.8 + 14.7 + 21.3 + 19.8}{5} = 16.36 \).
04

Compute the Overall Mean

Calculate the overall mean income using all the data points: \( \frac{12.7 + 9.2 + 10.9 + 8.9 + 16.4 + 8.3 + 17.2 + 19.1 + 10.3 + 20.3 + 16.6 + 22.7 + 25.2 + 19.9 + 17.2 + 8.8 + 14.7 + 21.3 + 19.8}{19} = 16.11 \).
05

Determine the ANOVA F-statistic

Use the formula for the ANOVA F-statistic: \[ F = \frac{(\text{Between-group variability})}{(\text{Within-group variability})} \].Calculate the variance for between-group and within-group as follows:- Between-group sum of squares (SSB): \( \frac{5(11.62 - 16.11)^2 + 4(13.725 - 16.11)^2 + 5(20.94 - 16.11)^2 + 5(16.36 - 16.11)^2}{3} \)- Within-group sum of squares (SSW):and degrees of freedom.
06

Compare with Critical Value

Find the critical value from the F-table for \( df_1 = 3 \) and \( df_2 = 15 \) at \( \alpha = 0.05 \). Compare the calculated F-statistic with the critical F-value.
07

Make a Decision

If the calculated F-statistic is greater than the critical F-value from the table, reject the null hypothesis. Otherwise, do not reject the null hypothesis.

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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 fundamental concept used in statistics to make informed conclusions about data. In the context of this example, we are using the ANOVA (Analysis of Variance) method. The goal is to test whether there is a significant difference in the income levels of immigrants from different countries.

The first step in hypothesis testing is to establish two mutually exclusive statements:
  • The **null hypothesis (H鈧)** suggests that there is no difference in the mean incomes across the groups. In our example, it states that the mean incomes of immigrants from the four countries are equal.
  • The **alternative hypothesis (H鈧)** proposes that at least one group's mean income differs from the others. This hypothesis suggests variability among the group means.
We use statistical tests, like ANOVA, to determine whether to reject the null hypothesis in favor of the alternative. If evidence strongly contradicts H鈧, we lean toward H鈧.

The outcome of hypothesis testing can tell whether observed differences are significant or if they could have occurred by random chance.
Level of Significance
The level of significance, denoted by the Greek letter alpha (伪), is a threshold set by the researcher. It determines the probability of rejecting the null hypothesis when it is actually true. In other words, it鈥檚 the risk of making a Type I error.

In our example, the level of significance is set at 0.05 or 5%. This implies a 5% chance of mistakenly concluding that there is a difference in income levels when, in reality, there isn't one.

Setting a significance level involves a trade-off: a lower 伪 reduces the risk of a Type I error but increases the risk of a Type II error, which is not detecting a difference when there is one. Common 伪 values include 0.05, 0.01, and 0.10, with 0.05 being the most widely used.

This benchmark helps in making informed decisions over the null hypothesis by comparing test results to this predetermined standard of uncertainty.
Mean Comparison
Mean comparison involves calculating and examining the average values for different groups to identify any disparities. In the case of ANOVA, we compute the means for each group separately to gain insights into their individual characteristics.

For our exercise, the mean incomes for the four groups are computed as follows:
  • Country I: 11.62
  • Country II: 13.725
  • Country III: 20.94
  • Country IV: 16.36
By comparing these values, we observe which groups have higher or lower average incomes, providing preliminary insights into group differences. Additionally, the overall mean, calculated as 16.11, serves as a reference point to assess individual group deviations.

The disparity in group means lays the groundwork for subsequent statistical testing. ANOVA specifically uses mean comparisons to examine if observed differences are statistically significant.
F-statistic Calculation
The F-statistic is a crucial figure used in ANOVA to determine variance among mean scores. It's a ratio that compares between-group and within-group variability. A higher F-value typically indicates greater disparity between group means relative to the variability within the groups.

To calculate the F-statistic:
  1. Compute the **Between-group sum of squares (SSB)**, which indicates the variance attributed to differences between group means.
  2. Calculate the **Within-group sum of squares (SSW)**, measuring variance within each group.
  3. Determine degrees of freedom for both between-group and within-group variances.
The formula for F is:
\[ F = \frac{(SSB / df_1)}{(SSW / df_2)} \]
Where \(df_1\) and \(df_2\) are the degrees of freedom.

This test statistic is compared against a critical value obtained from F-distribution tables, aiding in the decision to accept or reject the null hypothesis. If the calculated F exceeds the critical value, it signifies significant differences in group means, justifying rejection of H鈧.

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Most popular questions from this chapter

A civil engineer has been studying the frequency of vehicle accidents on a certain stretch of interstate highway. Longterm history indicates that there has been an average of \(1.72\) accidents per day on this section of the interstate. Let \(r\) be a random variable that represents number of accidents per day. Let \(O\) represent the number of observed accidents per day based on local highway patrol reports. A random sample of 90 days gave the following information. $$ \begin{array}{l|rrrrc} \hline \boldsymbol{r} & 0 & 1 & 2 & 3 & 4 \text { or more } \\ \hline 0 & 22 & 21 & 15 & 17 & 15 \\ \hline \end{array} $$ (a) The civil engineer wants to use a Poisson distribution to represent the probability of \(r\), the number of accidents per day. The Poisson distribution is $$ P(r)=\frac{e^{-\lambda} \lambda^{r}}{r !} $$ where \(\lambda=1.72\) is the average number of accidents per day. Compute \(P(r)\) for \(r=0,1,2,3\), and 4 or more. (b) Compute the expected number of accidents \(E=90 P(r)\) for \(r=0,1,2,3\), and 4 or more. (c) Compute the sample statistic \(\chi^{2}=\Sigma \frac{(O-E)^{2}}{E}\) and the degrees of freedom. (d) Test the statement that the Poisson distribution fits the sample data. Use a \(1 \%\) level of significance.

When the sample evidence is sufficient to justify rejecting the null hypothesis in a goodness-of-fit test, can you tell exactly how the distribution of observed values over the specified categories differs from the expected distribution? Explain.

A new fuel injection system has been engineered for pickup trucks. The new system and the old system both produce about the same average miles per gallon. However, engineers question which system (old or new) will give better consistency in fuel consumption (miles per gallon) under a variety of driving conditions. A random sample of 31 trucks were fitted with the new fuel injection system and driven under different conditions. For these trucks, the sample variance of gasoline consumption was 58.4. Another random sample of 25 trucks were fitted with the old fuel injection system and driven under a variety of different conditions. For these trucks, the sample variance of gasoline consumption was \(31.6 .\) Test the claim that there is a difference in population variance of gasoline consumption for the two injection systems. Use a \(5 \%\) level of significance. How could your test conclusion relate to the question regarding the consistency of fuel consumption for the two fuel injection systems?

Wild irises are beautiful flowers found throughout the United States, Canada, and northern Europe. This problem concerns the length of the sepal (leaf-like part covering the flower) of different species of wild iris. Data are based on information taken from an article by \(\mathrm{R} .\) A. Fisher in Annals of Eugenics (Vol. 7, part 2, pp. 179-188). Measurements of sepal length in centimeters from random samples of Iris setosa (I), Iris versicolor (II), and Iris virginica (III) are as follows: \(\begin{array}{ccc}\text { I } & \text { II } & \text { III } \\ 5.4 & 5.5 & 6.3 \\ 4.9 & 6.5 & 5.8 \\ 5.0 & 6.3 & 4.9 \\ 5.4 & 4.9 & 7.2 \\ 4.4 & 5.2 & 5.7 \\ 5.8 & 6.7 & 6.4\end{array}\) \(5.5\) Shall we reject or not reject the claim that there are no differences among the population means of sepal length for the different species of iris? Use a \(5 \%\) level of significance.

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