Question

Property Values In Chapter 10 , you compared the property evaluations of two tax assessors. \(A\) and \(B\), as shown in the table below. $$ \begin{array}{ccc} \hline \text { Property } & \text { Assessor } \mathrm{A} & \text { Assessor B } \\ \hline 1 & 276.3 & 275.1 \\ 2 & 288.4 & 286.8 \\ 3 & 280.2 & 277.3 \\ 4 & 294.7 & 290.6 \\ 5 & 268.7 & 269.1 \\ 6 & 282.8 & 281.0 \\ 7 & 276.1 & 275.3 \\ 8 & 279.0 & 279.1 \\ \hline \end{array} $$ a. Use the sign test to determine whether the data present sufficient evidence to indicate that assessor A tends to give higher assessments than assessor B; that is, \(P\left(x_{\mathrm{A}}\right.\) exceeds \(\left.x_{\mathrm{B}}\right)>1 / 2\). Test by using a value of \(\alpha\) near. \(05 .\) Find the \(p\) -value for the test and interpret its value. b. In Chapter 10 , we used the \(t\) statistic to test the null hypothesis that assessor A tends to give higher assessments than assessor \(\mathrm{B}\), resulting in a \(t\) -value of \(t=2.82\) with \(p\) -value \(=.013 .\) Do these test results agree with the results in part a? Explain why the answers are (or are not) consistent.

Short answer

Based on the results obtained using the sign test, we cannot conclude that assessor A tends to give higher assessments than assessor B, as the p-value is greater than the chosen significance level. However, the p-value obtained in the t-test from Chapter 10 suggests that there is a significant difference between the assessments of the two assessors. The discrepancy between the results can be attributed to the difference in statistical tests used. The sign test is less powerful and only considers the direction of the difference, while the t-test considers the magnitude of the difference as well.

Step-by-step solution

Compare the Assessments of A and B

First, we will compare the assessments given by assessor A and B for each property and note the number of times assessor A gives higher evaluations than assessor B. Property 1: A > B Property 2: A > B Property 3: A > B Property 4: A > B Property 5: A < B Property 6: A > B Property 7: A > B Property 8: A < B Excluding the cases in which both assessors give the same assessment, the total number of cases where A > B is 6.

Calculate p-value Using Binomial Test

Using the binomial test, we will find the probability of observing 6 or more successes in 8 trials, assuming the null hypothesis is true (\(P(a>b)=\frac{1}{2}\)). The binomial test formula is given by: $$p(x) = \binom{n}{x}p^x(1-p)^{n-x}$$ Now, let's plug our values and calculate the p-value. $$p(\geq 6) = \sum_{x=6}^8 \binom{8}{x} \left(\frac{1}{2}\right)^x \left(1 - \frac{1}{2}\right)^{8-x}$$ $$p(\geq 6) = \binom{8}{6} \left(\frac{1}{2}\right)^6 \left(\frac{1}{2}\right)^{2} + \binom{8}{7} \left(\frac{1}{2}\right)^7 \left(\frac{1}{2}\right)^{1} + \binom{8}{8} \left(\frac{1}{2}\right)^8 \left(\frac{1}{2}\right)^{0}$$ Calculate the p-value: $$p(\geq 6) = 28\times \frac{1}{64} + 8\times \frac{1}{128} + 1\times \frac{1}{256} = \frac{28}{64} + \frac{8}{128} + \frac{1}{256} = 0.0898$$

Interpret the p-value

The p-value obtained in part a is 0.0898, which is greater than the chosen significance level of \(\alpha = 0.05\). This means that we do not have sufficient evidence to reject the null hypothesis that assessor A tends to give higher assessments than assessor B, in favor of the alternative hypothesis.

Compare p-values of Both Tests

In part b, we are given the results of a t-test performed in Chapter 10, which resulted in a t-value of \(2.82\) and a p-value of \(0.013\). The p-value obtained in part a (0.0898) is greater than the p-value obtained in the t-test (0.013). The results are not consistent, which can be attributed to the different statistical tests used in each case. The sign test is less powerful than the t-test because it only considers the direction of the difference, not the magnitude of the difference.

Key concepts

Probability and Statistics

Understanding probability and statistics is fundamental for interpreting data and making decisions based on that data. Probability is the measure of the likelihood that an event will occur, and statistics is the science of collecting, analyzing, interpreting, and presenting data.

In the context of comparing property valuations by two tax assessors, we can use statistical methods to determine if there is a significant difference in their assessments. The data must be collected and organized effectively, which is often represented in a tabular form. From there, statistical tests are applied to draw conclusions about the population from which the sample is taken.

In this case, the sign test is used to compare the paired differences in assessments without making assumptions about the distribution of these differences—it's a form of non-parametric statistic. Its simplicity makes it appealing, but it may not be as powerful or informative as other statistical tests that take more information into account.

Binomial Test

The binomial test is a specific type of hypothesis test that can determine whether the number of successes in a sequence of yes-or-no experiments matches with a specified distribution. In our context, the 'success' is defined as an event where assessor A's valuation is higher than assessor B's.

The binomial formula \(p(x) = \binom{n}{x}p^x(1-p)^{n-x}\) accounts for the number of ways to achieve exactly \(x\) successes in \(n\) trials, with a constant probability \(p\) for each trial. When applied to the example of property assessments, it allows us to calculate the probability of observing a certain number of properties where A's assessment is higher than B's, under the null hypothesis that there is no difference between assessors' evaluations (with \(p = 0.5\)). The result of this calculation is a p-value that aids in making decisions about the null hypothesis.

p-value

The p-value is a critical concept in hypothesis testing. It indicates the probability of obtaining test results at least as extreme as the ones observed during the test, under the assumption that the null hypothesis is correct.

In simpler terms, the p-value helps us decide whether to reject the null hypothesis. If the p-value is less than or equal to the significance level \(\alpha\), usually set at 0.05, we reject the null hypothesis and accept that our results are statistically significant. In the exercise, the calculated p-value was 0.0898, which is higher than the standard 0.05. Therefore, we do not have enough evidence to conclude that assessor A consistently gives higher valuations than assessor B. It's essential to correctly interpret the p-value, as a lower p-value means stronger evidence against the null hypothesis, not a measure of effect size or importance of the results.

t-test

A t-test is another commonly used hypothesis test that compares two averages (means) and tells you if they are different from each other. The t-test also tells you how significant the differences are; in other words, it lets you know if those differences could have happened by chance.

In the property valuation exercise, a t-test was previously conducted, resulting in a t-value of 2.82 and a p-value of 0.013. This p-value is significantly below the 0.05 threshold, suggesting strong evidence against the null hypothesis and implying that assessor A tends to give higher valuations than assessor B. The discrepancy between the p-values obtained from the different tests highlights that while both tests aim to measure the same hypothesis, they might produce different results due to their sensitivity and the types of data they analyze.

Importantly, the t-test assumes that the data follow a normal distribution and measures the magnitude of differences, not just the direction, making it more powerful than the sign test in detecting significant differences in many cases.