/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 A particular county employs thre... [FREE SOLUTION] | 91Ó°ÊÓ

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A particular county employs three assessors who are responsible for determining the value of residential property in the county. To see whether these assessors differ systematically in their assessments, 5 houses are selected, and each assessor is asked to determine the market value of each house. With factor \(A\) denoting assessors \((I=3)\) and factor \(B\) denoting houses \((J=5)\), suppose \(\mathrm{SSA}=11.7, \mathrm{SSB}=113.5\), and \(\mathrm{SSE}\) \(=25.6\). a. Test \(H_{0}: \alpha_{1}=\alpha_{2}=\alpha_{3}=0\) at level .05. ( \(H_{0}\) states that there are no systematic differences among assessors.) b. Explain why a randomized block experiment with only 5 houses was used rather than a one-way ANOVA experiment involving a total of 15 different houses with each assessor asked to assess 5 different houses (a different group of 5 for each assessor).

Short Answer

Expert verified
a. Fail to reject \(H_0\); no systematic differences among assessors. b. Randomized block design controls for house variability.

Step by step solution

01

Understand the Hypotheses

The null hypothesis is \(H_0: \alpha_1 = \alpha_2 = \alpha_3 = 0\), suggesting no systematic differences among assessors. The alternative hypothesis is \(H_1: \alpha_i eq 0\) for at least one \(i\), indicating there is a difference among assessors.
02

Calculate Total Sum of Squares (SST)

The total sum of squares (SST) is calculated as the sum of SSA, SSB, and SSE. Thus, \(\text{SST} = \text{SSA} + \text{SSB} + \text{SSE} = 11.7 + 113.5 + 25.6 = 150.8\).
03

Calculate Mean Squares

Compute Mean Square for Assessors (\(MSA\)) and Mean Square for Error (\(MSE\)).\\(MSA = \frac{SSA}{I-1} = \frac{11.7}{3-1} = 5.85\) and \(MSE = \frac{SSE}{(I-1)(J-1)} = \frac{25.6}{(3-1)(5-1)} = 3.2\).
04

Compute the F-statistic

The F-statistic for testing \(H_0\) is calculated as \(F = \frac{MSA}{MSE} = \frac{5.85}{3.2} \approx 1.828\).
05

Determine the Critical Value and Decision

Use the F-distribution table to find the critical F value for \(\alpha = 0.05\), \(df_1 = 2\) (numerator) and \(df_2 = 8\) (denominator). The critical F value is approximately 4.46. Since the calculated F (1.828) is less than 4.46, we fail to reject \(H_0\).
06

Explain the Randomized Block Design

A randomized block design with only 5 houses is used to control for variability among houses. This design compares assessors for the same houses, reducing the impact of house-specific differences, whereas a one-way ANOVA would mix variability from different houses.

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Key Concepts

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

Randomized Block Design
The Randomized Block Design is a powerful statistical tool used to manage variability in experiments. In the context of this exercise, this design was applied to compare the performance of three different assessors evaluating the same set of five houses. By doing so, the focus remains strictly on the variability in assessments attributable to the assessors themselves, rather than differences between the houses.

This approach effectively "blocks" out the variation caused by different house characteristics, ensuring that the comparison among the assessors is fair and unbiased. Each assessor provides an evaluation of the same group of houses, allowing the differences in scores to be directly attributed to the assessors' individual judgment and expertise rather than external factors.

The randomized block design is ideal in situations where experimental units (in this case, houses) have some inherent variability. By controlling for this, researchers can obtain more precise estimates of treatment effects, leading to more valid conclusions.
F-statistic
The F-statistic is crucial in analysis of variance (ANOVA) for testing hypotheses about group means. It helps in determining whether any significant differences exist among the group means being compared. In this exercise, the F-statistic was calculated to assess whether differences in assessments by the three assessors could be statistically significant.

To compute the F-statistic, the mean square for the assessors (MSA) and the mean square for error (MSE) are utilized. The formula is \(F = \frac{MSA}{MSE}\), which, in this particular instance, resulted in an approximate value of 1.828.

This statistical measure essentially informs us of how much the variability between the group's mean assessments is greater than the variability within the assessments themselves. A larger F-statistic value indicates that the group means are significantly different from one another, whereas smaller values suggest that any observed differences may just be due to random chance.
Critical Value
In hypothesis testing, the critical value acts as a boundary for decision-making. It helps determine whether to reject the null hypothesis. For this exercise, the critical value is based on an F-distribution with parameters that depend on the degrees of freedom for the numerator (df_1) and denominator (df_2).

At a significance level of \(\alpha = 0.05\), the critical F value here is approximately 4.46. This means that for differences among assessors to be statistically significant, our calculated F value must exceed 4.46.

In cases where the computed F-statistic (1.828) is less than the critical value, the conclusion is to fail to reject the null hypothesis, indicating no sufficient evidence of significant differences among the assessors' evaluations.
Mean Squares
Mean squares are a key component in the ANOVA process. They represent the average of squared deviations, giving a sense of variability within and between groups.

In this exercise, we calculated two types of mean squares:
  • Mean Square for Assessors ( MSA): This is derived by dividing the Sum of Squares for Assessors ( SSA) by the degrees of freedom ( I-1), resulting in a value that measures variability between assessor evaluations.
  • Mean Square for Error ( MSE): Calculated by dividing the Sum of Squares for Error ( SSE) by the degrees of freedom ( (I-1)(J-1)), it represents the inherent variability within the evaluations not attributed to assessors.

MSA and MSE are key to computing the F-statistic, which is instrumental in hypothesis testing. The calculation and comparison of these mean squares provide insights into the sources and magnitude of variability in experimental data.

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