SOLUTION: A real estate agent wishes to determine whether tax assessors and real estate appraisers agree on the value of homes. A random sample of the two groups appraised 10 homes. The data

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Question 1200994: A real estate agent wishes to determine whether tax assessors and real estate appraisers agree on the value of homes. A random sample of the two groups appraised 10 homes. The data are shown here. Is there a significant difference in the values of the homes for each group? Let a=0.05. Find the 95% confidence interval for the difference of the means.
Real estate appraisers Tax assessors
Xbar1 = $83,256 Xbar2 = $88,354
S1 = $3256 S2 = $2341
n1 = 10 n2 = 10
Answer by math_tutor2020(3817) About Me  (Show Source):
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We're doing a two-sample t-test.

Refer to these pages
https://stattrek.com/hypothesis-test/difference-in-means
https://www.statology.org/welchs-t-test/
for more information and other examples.
We will not be pooling the data because "In practice, when you are comparing the means of two groups it’s unlikely that the standard deviations for each group will be identical" (quote from the 2nd link).

Given information:
Xbar1 = $83,256 Xbar2 = $88,354
S1 = $3256 S2 = $2341
n1 = 10 n2 = 10

Compute the point estimate
Xbar1-Xbar2 = 83256-88354
Xbar1-Xbar2 = -5098
This estimates the value of mu1-mu2.

To determine the degrees of freedom (df), we pick the smaller of n1-1 and n2-1
But since n1 = n2, we basically compute the following: df = n-1 = 10-1 = 9
Then use a T table such as this one
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
to find the t critical value is roughly t = 2.262
Look at the row df = 9 and the column labeled "two tails 0.05" to locate t = 2.262
The "two tails 0.05" means 0.05 is the total area of both tails combined, meaning 1-0.05 = 0.95 = 95% of the area is located in the middle. This corresponds to a 95% confidence level.

Null Hypothesis:
H0: mu1 = mu2 aka mu1-mu2 = 0
Alternative Hypothesis
H1: mu1 =/= mu2 aka mu1-mu2 =/= 0

Compute the margin of error for the difference in means.
E+=+t%2Asqrt%28+%28%28s1%29%5E2%29%2F%28n1%29%2B%28%28s2%29%5E2%29%2F%28n2%29+%29+

E+=+2.262%2Asqrt%28+%28%283256%29%5E2%29%2F%2810%29%2B%28%282341%29%5E2%29%2F%2810%29+%29+

E+=+2868.53503485922

E+=+2868.535+

Now we can calculate the boundaries of the confidence interval.

L = lower boundary of the confidence interval
L = pointEstimate - marginOfError
L = (xbar1-xbar2) - E
L = -5098 - 2868.535
L = -7966.535

U = upper boundary of the confidence interval
U = pointEstimate + marginOfError
U = (xbar1-xbar2) + E
U = -5098 + 2868.535
U = -2229.465

The 95% confidence interval in the format
L < mu1 - mu2 < U
is approximately
-7966.535 < mu1 - mu2 < -2229.465

The difference in population means is between -7966.535 and -2229.465; we are 95% confident of this statement.

The value 0 is NOT in this interval, which means that mu1-mu2 = 0 is probably not likely.
Furthermore, it means mu1 = mu2 is probably not likely.

Therefore, we reject the null and conclude that
mu1 =/= mu2 aka mu1-mu2 =/= 0
is likely the case.

Conclusion: There appears to be a significant difference in the home values for each group.