In: Statistics and Probability
The city of Wenatchee employs people to assess the value of homes for the purpose of establishing real estate tax. The city manager sends each assessor to the same 5 homes and then compares the results. The information is given below, in thousands of dollars. Can we conclude that there is a difference in the assessors, at Alpha= .05
Assessor | ||||
Home | Livingston | Wilson | Eilers | Phelps |
A | $51 | $55 | $49 | $45 |
B | $50 | $45 | $52 | $53 |
C | $53 | $54 | $47 | $55 |
D | $70 | $72 | $62 | $64 |
E | $82 | $89 | $92 | $81 |
Is there a difference in the population means?
A) What test should be run?
B) State the null and alternate hypothesis
C) Compute the value of the test statistic
D) Calculate the P-Value
E) What is your decision regarding the null hypothesis? Interpret the result.
Is there a difference in block means?
A) State the null and alternate hypothesis
B) Compute the value of the test statistic
C) Calculate the P-Value
D) What is your decision regarding the null hypothesis? interpret the result.
For testing that there is a difference in the assessors and as well in different homes from the given data we can test the hypothesis by using two-way ANOVA model.
To test that is there any difference in population mean?
A) We should run F-test
B) H0: there is no difference in population mean Vs H1: there is difference between any two population means
C) the value of test statistics is = 51.97
D) P-value = 0.0001
E) since the P -value is smaller than alpha=0.05, we reject null hypothesis that, there is no difference in population means. and hence we can say that there may chance of having any two population means to be equal.
To test that, is there any difference in blocks
A) H0: there is no any difference between block means Vs H1: there is difference between any two blocks means.
B) The value of test statistics is = 31.8
C) P-value = 0.689
D) Since P-value is much larger that alpha=0.05, We fail to reject the null hypothesis and hence we should say that the block means are equal with 5% level of significance.