91Ó°ÊÓ

A neighborhood home owners association suspects that the recent appraisal values of the houses in the neighborhood conducted by the county government for taxation purposes is too high. It hired a private company to appraise the values of ten houses in the neighborhood. The results, in thousands of dollars, are \(\begin{array}{|l|c|c|} \hline \text { House } & \text { County Government } & \text { Private Company } \\ \hline 1 & 217 & 219 \\ \hline 2 & 350 & 338 \\ \hline 3 & 296 & 291 \\ \hline 4 & 237 & 237 \\ \hline \end{array}\) \(\begin{array}{|l|c|c|} \hline \text { House } & \text { County Government } & \text { Private Company } \\ \hline 5 & 237 & 235 \\ \hline 6 & 272 & 269 \\ \hline 7 & 257 & 239 \\ \hline 8 & 277 & 275 \\ \hline 9 & 312 & 320 \\ \hline 10 & 335 & 335 \\ \hline \end{array}\) a. Give a point estimate for the difference between the mean private appraisal of all such homes and the government appraisal of all such homes. b. Construct the \(99 \%\) confidence interval based on these data for the difference. c. Test, at the \(1 \%\) level of significance, the hypothesis that appraised values by the county government of all such houses is greater than the appraised values by the private appraisal company.

Step by step solution

01

Calculate the Differences

To find the difference between the appraisals of the county government and the private company for each house, subtract the private appraisal from the county appraisal. House 1: 217 - 219 = -2 House 2: 350 - 338 = 12 House 3: 296 - 291 = 5 House 4: 237 - 237 = 0 House 5: 237 - 235 = 2 House 6: 272 - 269 = 3 House 7: 257 - 239 = 18 House 8: 277 - 275 = 2 House 9: 312 - 320 = -8 House 10: 335 - 335 = 0 Thus, the differences are: -2, 12, 5, 0, 2, 3, 18, 2, -8, 0.

02

Compute the Mean Difference

Find the mean of the differences calculated in Step 1. Sum the differences and divide by the number of observations (10).Sum of differences: -2 + 12 + 5 + 0 + 2 + 3 + 18 + 2 - 8 + 0 = 32Mean difference: \[ \bar{d} = \frac{32}{10} = 3.2 \]

03

Calculate the Standard Deviation of Differences

First, calculate the squared differences from the mean, then find their average and take the square root.Squared differences: (-2-3.2)^2, (12-3.2)^2, ..., (0-3.2)^2These calculate to: 27.04, 77.44, 3.24, 10.24, 1.44, 0.04, 216.36, 1.44, 76.16, 10.24Sum of squared differences: 423.6Variance: \[ s^2 = \frac{423.6}{9} = 47.0667 \]Standard deviation: \[ s = \sqrt{47.0667} \approx 6.86 \]

04

Construct the 99% Confidence Interval

Use the formula for the confidence interval:\[ \text{CI} = \bar{d} \pm t \frac{s}{\sqrt{n}} \]For \( n = 10 \), degrees of freedom \( df = 9 \), and \( t = 3.250 \) for 99% confidence.\[ CI = 3.2 \pm 3.250 \times \frac{6.86}{\sqrt{10}} \approx 3.2 \pm 7.06 \]Thus, the 99% confidence interval is \(-3.86\) to \(10.26\).

05

Hypothesis Testing

To test if the government appraisals are higher:- Null hypothesis \( H_0: \mu_d = 0 \)- Alternative hypothesis \( H_a: \mu_d > 0 \)Use the test statistic:\[ t = \frac{\bar{d} - 0}{s / \sqrt{n}} = \frac{3.2}{6.86/\sqrt{10}} \approx 1.48 \]With \( t = 1.48 \), compare this with the critical \( t \) value of 3.250.Since 1.48 < 3.250, we fail to reject the null hypothesis.

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Key Concepts

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

Point Estimate

Point estimate refers to the single value that serves as our best guess for the parameter we're interested in estimating. In this exercise, we are looking for the point estimate of the difference between the mean private appraisal and the government appraisal of the homes in the neighborhood.
To find this, we first calculate the difference between the government's and private company's appraisals for each house. These differences help us understand if there's a significant disparity between the two appraisal methods. The differences for each house were calculated as -2, 12, 5, 0, 2, 3, 18, 2, -8, 0.
The mean of these differences gives us our point estimate for the average disparity in appraisals, which is calculated as follows:

• Sum of differences: -2 + 12 + 5 + 0 + 2 + 3 + 18 + 2 -8 + 0 = 32
• Number of houses: 10

Hence, the point estimate, or mean difference, is \( \bar{d} = \frac{32}{10} = 3.2 \). This tells us that, on average, the government's appraisal is 3.2 units more than the private company’s.

Confidence Interval

A confidence interval provides a range of values within which we expect the true parameter to fall with a certain level of confidence. In this scenario, the objective is to construct a 99% confidence interval for the difference between the appraisals.
The calculation involves the sample mean difference, the standard deviation of differences, and the appropriate t-value. For a 99% confidence level with 9 degrees of freedom (since there are 10 houses, giving us 9 degrees of freedom), we use a t-value of 3.250.
The formula for the confidence interval is:

• \( \text{CI} = \bar{d} \pm t \frac{s}{\sqrt{n}} \)

Substitute the values into the formula:

• Mean difference, \( \bar{d} = 3.2 \)
• Standard deviation, \( s = 6.86 \)
• t-value = 3.250
• Number of observations, \( n = 10 \)

The confidence interval calculation becomes:
\[ CI = 3.2 \pm 3.250 \times \frac{6.86}{\sqrt{10}} \approx 3.2 \pm 7.06 \]
Thus, the 99% confidence interval is from -3.86 to 10.26. This interval indicates that we're 99% confident the true difference in appraisals falls within this range.

Standard Deviation

Standard deviation measures the amount of variation or dispersion in a set of values. It informs us about how spread out the differences are from the mean difference.
To calculate it, first find the squared differences from the mean. Then calculate their mean and take the square root, which gives us the standard deviation of the differences.

• Differences: -2, 12, 5, 0, 2, 3, 18, 2, -8, 0
• Mean difference, \( \bar{d} = 3.2 \)

Next, calculate the squared differences:

• \((-2-3.2)^2, (12-3.2)^2, ..., (0-3.2)^2\)
• This results in: 27.04, 77.44, 3.24, 10.24, 1.44, 0.04, 216.36, 1.44, 76.16, 10.24
• Sum these squared differences: 423.6
• Then compute the variance: \( s^2 = \frac{423.6}{9} = 47.0667 \)
• The standard deviation \( s = \sqrt{47.0667} \approx 6.86 \)

The standard deviation of 6.86 indicates there's a fair amount of variability in the differences,

Null Hypothesis

The concept of null hypothesis is central to hypothesis testing. It represents the default position that there is no effect or difference.
In this exercise, the null hypothesis is \( H_0: \mu_d = 0 \), which implies there is no difference between the government and private appraisals. The alternative hypothesis \( H_a: \mu_d > 0 \) suggests that the government's appraisals are higher.
We test this using a t-test. Here, the test statistic is:
\[ t = \frac{\bar{d} - 0}{s / \sqrt{n}} = \frac{3.2}{6.86/\sqrt{10}} \approx 1.48 \]

The critical value for a 99% confidence level and 9 degrees of freedom is 3.250. Since our calculated t-value of 1.48 is less than 3.250, we fail to reject the null hypothesis. This means the data doesn't provide enough evidence to conclude that government appraisals are significantly higher than private appraisals.