The city of Wenatchee, Washignton, emplys people to asses the value of homes for

Question

The city of Wenatchee, Washignton, emplys people to asses the value of homes for the purpose of extablishing real estate tax. The city manager send each assessor to the same five homes and then compares the results. The information is given below, in thousnds of dollars. Can we conclude that there is a diffrent in the assesors, at a=0.05?

Assesors

Is there a diffrence in the populatin mean?

a.What test should I run?

b.State the null and alternate hypotheses

c. find the critical value

d.Compute the value test statistic

e. What is your decision rule regarding the null hypothesis? Interpret the result

Is there a diffrence in the block?

f. Sate the null phypothesis

g.find the critical value

h.compute the value of test statistic

i. What is your decision regarding the null hypotheis?Intepret resuts

Home Livingston Wilson Eilers Phels A $49 $55 $49 $45 B $50 $45 $52 $53 C $50 $52 $47 $53 D $70 $72 $62 $64 E $84 $89 $92 $87

Explanation / Answer

a) ANOVA one way analysis

b) Null hypothesis : There is no significance difference between all the asses

Alternate hypothesis : There is significance difference between all the asses

c) Critical value

degrees of freedom (3,16) at 0.05 level of F test

= 3.238875

Q1= SS - T^2/n

= 79146 - 1220^2/20

= 4726

Q2 = Ti^2/N - T^2/n

= 74437.2 - 1220^2/20

= 17.2

Since the P value is 0.991615 > 0.05, there is no significance difference between grps

Reject alternate hypothesis

_________________________________________________________________________________

f) Null : There is no difference in houses

ALternate : There is differece in houses

critical value = 3.055

Accept alternate hypothesis

Home Livingston Wilson Eilers Phels A 49 55 49 45 B 50 45 52 53 C 50 52 47 53 D 70 72 62 64 E 84 89 92 87 Sum(T) 303 313 302 302 1220 T^2/4 18361.8 19593.8 18240.8 18240.8 74437.2 SS 19357 20859 19622 19308 79146

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