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StatisticsQ&A Library9.7 Refer to Example 8.6. The political action group was interested in determining regionaldifferences in the public's opinion concerning air pollution. Write a contrast in the four populationmeans to answer each of the following questionsa. Question 1: Is the proportion of people who thought the EPA's standards arenot stringent enough different for the people living in the East compared to thepeople living in the West?b. Question 2: Is the proportion of people who thought the EPA's standards are notstringent enough different for the people living in the Northeast compared to thepeople living in the other three regions?c. Question 3: Is the proportion of people who thought the EPA's standards are notstringent enough different for the people living in the Northeast compared to thepeople living in the Southeast?d. Simultaneously test if the three contrasts are different from 0 using an αe. Are the three contrasts mutually orthogonal?.05 test.Refer to Example 8.6 Data setRegionSMSANESE.817810.12.13.605656MeanStandard Deviation343.02730274.0179Question

Asked Mar 18, 2019

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Step 1

**Note:**

Hey there! Thank you for the question. Since your question consists of at least 3 different questions, we have solved the first question (Question 1 in part (a)) in detail for you and given an outline as to how the other parts can be solved in a similar manner. If you need help with any other part, then please re-post the question and specify the part you need answered.

Step 2

**Part (a): Question 1:**

**Contrast and Sum of Square of Contrast:**

A contrast is defined as a linear combination of *g* treatment means *μ _{i}*, (for

Denote *μ*_{1}, *μ*_{2},* μ*_{3} and *μ*_{4} as the population mean proportions of people in the Northeast (NE), Southeast (SE), Midwest (MW) and West (W) respectively, who thought the EPA’s standards concerning air pollution were not stringent enough.

Now, East (E) comprises of NE and SE; W comprises of MW and W.

The contrast in the 4 population means, in order to determine whether the proportion of people who thought the EPA’s standards concerning air pollution were not stringent enough is different for people in the East as compared to those in the West is:

*C*_{1}:* μ*_{1} + *μ*_{2} – (*μ*_{3} + *μ*_{4}) =* μ*_{1} + *μ*_{2} – *μ*_{3} – *μ*_{4}.

Here, *w*_{1} = *w*_{2} = 1; *w*_{3} = *w*_{4} = –1, so that ∑*w _{i}* = 0, when the sum is done over all

Denote the observations as *y _{ij}*, for the

The hypothesis to be tested is: *H*_{0} : ∑*w _{i} μ_{i}* = 0, vs,

The sum of square of the contrast is:

Step 3

**Mean square error:**

Let *Si*2 be the within treatment sum of squares for the *i*th treatment (in this case, the regions), taken about the treatment mean.

In case the *i*th treatment has *ni* observations and the sample variance for the *i*th treatment is known to be *Vi*, then, *Si*2 = (*ni* – 1)* Vi* [as sample variance is obtained by dividing the sum of squares by (*ni* – 1) instead of *ni*].

The, the error sum of squares can be obtained as: *SSE* = ∑*ni Si*2, when the sum is done over all *i* = 1, 2,3 4.

The mean square error is o...

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