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A well-known automotive magazine took three top-of-the-line midsize automobiles manufactured in the United States, test-drove them, and compared them on a variety of criteria. In the area of gasoline mileage performance, five automobiles of each brand were each test-driven 500 miles; the miles per gallon data obtained follow. Use \(\alpha=.05\) to test whether there is a significant difference in the mean number of miles per gallon for the three types of automobiles. $$\begin{array}{ccc} & \text { Automobile } & \\ \mathbf{A} & \mathbf{B} & \mathbf{C} \\ 19 & 19 & 24 \\ 21 & 20 & 26 \\ 20 & 22 & 23 \\ 19 & 21 & 25 \\ 21 & 23 & 27 \end{array}$$

Short Answer

Expert verified
There is a significant difference in the mean miles per gallon among the three automobile brands.

Step by step solution

01

Define the Hypotheses

We want to determine if there is a significant difference in the mean miles per gallon for the three types of automobiles. - Null Hypothesis (H_0): \( \mu_A = \mu_B = \mu_C \), meaning the means for all brands are equal.- Alternative Hypothesis (H_a): At least one pair of means is different.
02

Calculate the Means

Compute the mean miles per gallon for each automobile brand:\[\text{Mean for A: } \frac{19 + 21 + 20 + 19 + 21}{5} = 20 \]\[\text{Mean for B: } \frac{19 + 20 + 22 + 21 + 23}{5} = 21 \]\[\text{Mean for C: } \frac{24 + 26 + 23 + 25 + 27}{5} = 25 \]The means are 20, 21, and 25 for A, B, and C respectively.
03

Calculate the Overall Mean

The overall mean miles per gallon is calculated by averaging all the data:\[\text{Overall Mean: } \frac{19 + 21 + 20 + 19 + 21 + 19 + 20 + 22 + 21 + 23 + 24 + 26 + 23 + 25 + 27}{15} = 22 \]
04

Calculate the Sum of Squares

Compute the sum of squares for between-groups (SSB) and within-groups (SSW).- SSB:\[SSB = 5((20-22)^2 + (21-22)^2 + (25-22)^2) = 80 \]- SSW: Calculate SSW for each group and add:\[SSW = \sum (x - \bar{x})^2 \]\[= ((19-20)^2 + (21-20)^2 + ... + (27-25)^2) = 34 \]
05

Calculate the Mean Squares

Use the sum of squares to calculate the mean squares for both between and within groups:- Mean Square Between (MSB): \[MSB = \frac{SSB}{k-1} = \frac{80}{2} = 40 \]- Mean Square Within (MSW): \[MSW = \frac{SSW}{N-k} = \frac{34}{12} \approx 2.83 \]
06

Perform ANOVA F-test

Calculate the F statistic using the calculated mean squares:\[F = \frac{MSB}{MSW} = \frac{40}{2.83} \approx 14.14 \]Determine the critical value for \alpha = 0.05 with df1 = 2 and df2 = 12 from the F-distribution table and compare it to the calculated F value.
07

Decision and Conclusion

If the calculated F value is greater than the critical F value from the table, reject the null hypothesis. The critical F value is approximately 3.88. Since 14.14 > 3.88, we 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
In hypothesis testing, we are trying to determine if a specific statement about a population parameter is true. In this exercise, we are comparing the mean gasoline mileage for three different brands of automobiles.

The initial step involves setting up a Null Hypothesis (\( H_0 \)) and an Alternative Hypothesis (\( H_a \)).
  • The Null Hypothesis (\( H_0 \)) states that there is no difference in the mean miles per gallon for all automobile brands. In mathematical terms: \( \mu_A = \mu_B = \mu_C \).
  • The Alternative Hypothesis (\( H_a \)) claims that at least one brand has a different mean miles per gallon. This suggests that not all means are equal.
Hypothesis testing is crucial to determine if observed differences are due to sampling variability or if there are actual differences among groups.
Sum of Squares
The Sum of Squares (SS) helps us understand the variability present in our data. Within ANOVA, there are two main types of sum of squares: Between-groups Sum of Squares (SSB) and Within-groups Sum of Squares (SSW).

  • **Between-groups Sum of Squares (SSB):** This measures the variability due to the differences between group means and the overall mean. It tells us how much the group means deviate from the grand mean. In this problem: \( SSB = \sum n((\bar{x}_i - \bar{x})^2) \), which calculates to 80.
  • **Within-groups Sum of Squares (SSW):** This measures the variability within each group. It reflects how scores within each group differ from their group mean. For this exercise, \( SSW = \sum (x - \bar{x})^2 \) sums up to 34.
Understanding sum of squares is essential to analyze how much of the total variability can be attributed to differences within groups versus differences between groups.
Mean Squares
Mean Squares (MS) are derived from sums of squares, and they represent the average variability per degree of freedom. They are a vital part of ANOVA calculations.

  • **Mean Square Between (MSB):** This is calculated by dividing the between-groups sum of squares (SSB) by the degrees of freedom for the groups. Here, \( MSB = \frac{SSB}{k-1} = \frac{80}{2} = 40 \).
  • **Mean Square Within (MSW):** Similarly, this is computed by dividing the within-groups sum of squares (SSW) by its degrees of freedom. In this case, \( MSW = \frac{SSW}{N-k} = \frac{34}{12} \approx 2.83 \).
Mean squares help us understand the variation at each level of analysis, setting the stage for further hypothesis testing with the F-test.
F-test
The F-test is a statistical test used in ANOVA to determine if there are any statistically significant differences between the means of three or more groups. It compares the variability between the groups to the variability within groups.

To perform the F-test, we calculate the F statistic:
  • **F-statistic Formula:** This is given by \( F = \frac{MSB}{MSW} \). For this exercise, it becomes \( F = \frac{40}{2.83} \approx 14.14 \).
Once the F value is computed, it is compared against a critical value from the F-distribution table at a given significance level (\( \alpha \)).
  • The degrees of freedom for the numerator (between-groups) are \( k-1 \) and for the denominator (within-groups) are \( N-k \).
If the calculated F statistic is greater than the critical value, we reject the null hypothesis, suggesting at least one group mean is significantly different. Here, since \( 14.14 > 3.88 \), the null hypothesis is rejected, indicating a significant difference in gasoline mileage among the automobile brands.

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Most popular questions from this chapter

The following results are for independent random samples taken from two populations. $$\begin{array}{ll} \text { Sample 1 } & \text { Sample 2 } \\ n_{1}=20 & n_{2}=30 \\ \bar{x}_{1}=22.5 & \bar{x}_{2}=20.1 \\ s_{1}=2.5 & s_{2}=4.8 \end{array}$$ a. What is the point estimate of the difference between the two population means? b. What is the degrees of freedom for the \(t\) distribution? c. \(\quad\) At \(95 \%\) confidence, what is the margin of error? d. What is the \(95 \%\) confidence interval for the difference between the two population means?

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The U.S. Department of Transportation provides the number of miles that residents of the 75 largest metropolitan areas travel per day in a car. Suppose that for a simple random sample of 50 Buffalo residents the mean is 22.5 miles a day and the standard deviation is 8.4 miles a day, and for an independent simple random sample of 40 Boston residents the mean is 18.6 miles a day and the standard deviation is 7.4 miles a day. a. What is the point estimate of the difference between the mean number of miles that Buffalo residents travel per day and the mean number of miles that Boston residents travel per day? b. What is the \(95 \%\) confidence interval for the difference between the two population means?

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