Question

Please show work :)

6. Number of Farms The numbers in thousands) of farms per state found in three sections of the country is listed next. Test the claim at a a = 0:05 that the mean number of farms is the same across these three geographic divisions. Easter third 48 57 24 10 Middle third 95 52 64 64 Western third 29 40 40 68

Answer 1

Solution-

To test the claim that the mean number of farms is same across this three geographical divisions

to test above claim ANOVA test is calculated.

◆ Test Calculation and results-

Easter third Middle third Western third 48 95 29 57 52 40 24 64 40 10 64 68 Sum = 139 275 177 Average = 34.75 68.75 44.25 X = 6229 19921 8665 St. Dev. = 21.593 18.392 16.661 SS = 1398.75 1014.75 832.75 n = 4 4 4

The total sample size is N = 12. Therefore, the total degrees of freedom are: dftotal = 12 – 1= 11 3 – 1= 2, and the within-groups Also, the between-groups degrees of freedom are dfbetween degrees of freedom are: dfwithin = d ftotal - d fbetween 11 – 2 = 9 First, we need to compute the total sum of values and the grand mean. The following is obtained Xij = 139 + 275 + 177 = 591 i,j Also, the sum of squared values is x}; = 6229 + 19921 + 8665 = 34815 Based on the above calculations, the total sum of squares is computed as follows SStotal = x; = 34815 5912 12 5708.25 The within sum of squares is computed as shown in the calculation below: SSwithin = SSwithingroups 1398.75 + 1014.75 + 832.75 = 3246.25 The between sum of squares is computed directly as shown in the calculation below: The between sum of squares is computed directly as shown in the calculation below: Now that sum of squares are computed, we can proceed with computing the mean sum of squares: 2462 M Sbetween SSbetween dfbetween = 1231 2 SSwithin 3246.25 MS within 360.694 dfwithin 9 Now that sum of squares are computed, we can proceed with computing the mean sum of squares: 2462 M Sbetween SSbetween dfbetween 1231 2 SSwithin 3246.25 MSwithin 360.694 dfwithin 9 Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows: MSbetween F 1231 360.694 = 3.413 M Swithin

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: M1 = H2 = 43 Ha: Not all means are equal The above hypotheses will be tested using an F-ratio for a One-Way ANOVA. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are dfi 2 and df2 = 2, therefore, the rejection region for this F-test is R={F:F> F. = 256} (3) Test Statistics F MSbetween M Swithin 1231 360.694 3.413 (4) Decision about the null hypothesis Since it is observed that F = 3.413 <F. = 4.256, it is then concluded that the null hypothesis is not rejected. 0.0789 > 0.05, it is Using the P-value approach: The p-value is p = 0.0789, and since p concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that not all 3 population means are equal, at the a = 0.05 significance level.

by the above results null hypothesis Ho is not rejected.

So, we have enough evidence to support the claim that the mean number of farms is same across this three geographic divisions.