Question

1. A nonprofit consumer-protection organization compares three types of front bumpers for automobiles.   The organization wishes to see whether the type of bumper makes a difference in protecting cars from damage. Tests are conducted by driving an automobile into a concrete wall at 15 miles per hour. The outcome examined is the amount of damage to the car, as measured by repair costs in hundreds of dollars. Due to the large costs of this type of testing, the organization only conducts two tests with each bumper type. The following table shows the results. The organization decides to analyze the data by conducting an analysis of variance.

Bumper A	Bumper B	Bumper C
1	2	11
3	4	15

1. Write out the null and alternative hypotheses.
2. Find the within-groups sum of squares, its degrees of freedom, and the associated mean square.
3. Find the between-groups sum of squares, its degrees of freedom, and the associated mean square.
4. Calculate the test statistic and determine its p-value. Can you reject the null hypothesis at the 0.05 level? At the 0.01 level?

Answer 1

1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ₁ = μ₂ = μ₃

Ha: Not all means are equal.

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Between Groups Degrees of Freedom: DF = k − 1 , where k is the number of groups

Within Groups Degrees of Freedom: DF = N − k , where N is the total number of subjects

Total Degrees of Freedom: DF = N − 1

Sum of Squares Between Groups :

, where n_i is the number of subjects in the i-th group

Sum of Squares Within Groups:

, where S_i is the standard deviation of the i-th group

Total Sum of Squares:

Mean Square Between Groups:

Mean Square Within Groups:

F-Statistic (or F-ratio):

  

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Data summary :

Group	N	Mean	Std.Dev	Std.Error
Group 1	2	2	1.4142	1
Group 2	2	3	1.4142	1
Group 3	2	13	2.8284	2

degrees of freedom for SSw = n-k= 3

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c)  

degrees of freedom for SSb= k-1 = 3-1 = 2

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c)

To find p- value we use excel command as, =FDIST(f test statitics ,df for between , df for within , ) ==> FDIST(18.5004,2,3,)

You will get p- value as ,0.0205

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P- value rule for rejection :

If p- value < significance level ; reject H0

If p- value > significance level ; fail to reject H0.

Here p- value =0.0205 < 0.05 ; reject H0

We concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that not all 3 population means are equal, at the α=0.05 significance level

Here p- value 0.0205 > 0.01 ; fail to reject H0.

We concluded that the null hypothesis Ho is not  rejected. Therefore, there is enough evidence to claim that all 3 population means are equal, at the α=0.05 significance level