Question

I need help answering t test stastic equation .

2. After receiving the state tax assessment of $87,500 for his house, John decided to appeal this assessment. He was told that his assessment was made as the average of all houses in his neighborhood. John hired a consulting company to randomly sample and assess eight houses in the neighborhood. The results are: $85,200; $86,400; $84,250; $85,300; $87,600; $84,750; $90,100 and $85,800. The consultant verified that house values in John's neighborhood are normally distributed. Use a 0.05 level of significance and test the hypothesis that the average assessed value of all houses in John's neighborhood is less than the $87,500 he was originally assessed. Did John get the state to lower its tax assessment of his house?

Answer 1

From the given data, the calculation are as follows:

#	X	Mean	(x - mean)2
1	85200	86175	950625
2	86400	86175	50625
3	84250	86175	3705625
4	85300	86175	765625
5	87600	86175	2030625
6	84750	86175	2030625
7	90100	86175	15405625
8	85800	86175	140625

n	8
Sum	689400
Average	86175.00
SS	25080000
Variance = SS/n-1	3582857.143
Std Dev=Sqrt(Variance)	1892.84367

$\mu$ = $87,500, $\bar{x}$ = $86,175, s = $1892.843, n = 8

The Hypothesis:

H0: $\mu$ = $87,500 : The average assessed value in Johns neighbourhood is equal to $87,500.

Ha: $\mu$ < $87500 : The average assessed value in Johns neighbourhood is less than $87,500.

This is a left tailed test

The Test Statistic: Since the population standard deviation is unknown, we use the students t test.

The test statistic is given by the equation:

$t = \frac{\bar{x}-\mu }{\frac{s}{\sqrt{n}}} = \frac{86175-87500}{\frac{1892.843}{\sqrt{8}}} = -1.98$

t observed = -1.98

The p Value:    The p value (Left Tail) for t = -1.98, for degrees of freedom (df) = n-1 = 7, is; p value = 0.0440

The Critical Value:   The critical value (Left Tail) at $\alpha$ = 0.05, for df = 7, tcritical = -1.8946

The Decision Rule:   If tobserved is < -1.8946, Then Reject H0.

Also if P value is < $\alpha$ (0.05), Then Reject H0.

The Decision: Since tobserved (-1.98) is < -1.89 , We Reject H0.

Also since P value (0.0440) is < 0.05, We Reject H0.

The Conclusion: Reject H0. There is sufficient evidence at the 95% significance level to conclude that the average assessed value in Johns neighbourhood is less than $87,500.

Therefore Yes, John got the state to lower its tax assessment of his house.