A real estate agent wishes to determine whether a tax assessors and real estate appraises agree on the values 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 alpha = 0.05. Use p value method. Samplemean Real estate appraiser$83,256 tax assessors$88,354   s1= $3256s2= 2341  N1= 10n2= 10

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Asked Nov 25, 2019
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A real estate agent wishes to determine whether a tax assessors and real estate appraises agree on the values 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 alpha = 0.05. Use p value method. 

Sample

mean 

Real estate appraiser

$83,256 

tax assessors

$88,354 

 
  s1= $3256 s2= 2341  
  N1= 10 n2= 10  
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Expert Answer

Step 1

From the given information, the claim of the problem is there is a significance difference in the values of homes for real estate appraises and tax assessors.

 

Let us define µ1 be the population mean for real estate appraises and µ2 be the population mean for tax assessors.

 

Null hypothesis:

 

H0: µ1= µ2.

 

That is, there is no significance difference in the values of homes for real estate appraises and tax assessors.

 

Alternative hypothesis:

 

H1: µ1≠µ2.

 

That is, there is a significance difference in the values of homes for real estate appraises and tax assessors.

 

Step 2

From the given information, the sample mean for real estate appraises is $83,256, the sample mean for tax assessors is $88,354 , the sample standard deviation for real estate appraises is s1= 3256, the sample standard deviation for tax assessors is s2= 2341. The sample size for real estate appraises is n1=10 and the sample size for tax assessors is n2=10.

 

Here, the population standard deviations are unknown and the sample sizes are less than 30. Moreover, the two samples are independent. Hence, the suitable test for testing the given hypothesis is two-sample independent t test.

 

The test statistic is -4.02 and it is calculated below:

test
п,
п,
83,256 88,354
23412
32562
10
10
-5098
1268.141
=-4.02
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test п, п, 83,256 88,354 23412 32562 10 10 -5098 1268.141 =-4.02

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Step 3

The p-value is 0.0008 and it is calculated below:

 

The degrees of freedom is n1+n2...

p-value 2P(<t
-2P(t-4.02)
Since, Given Hypothesis is two tail test
test
0.000402=T.DIST(-4.02,8,TRUE)
-2 x0.000402
=0.0008
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p-value 2P(<t -2P(t-4.02) Since, Given Hypothesis is two tail test test 0.000402=T.DIST(-4.02,8,TRUE) -2 x0.000402 =0.0008

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