Question

Wind Mountain is an archaeological study area located in southwestern New Mexico. Potsherds are broken pieces of prehistoric Native American clay vessels. One type of painted ceramic vessel is called Mimbres classic black-on-white. At three different sites, the number of such sherds was counted in local dwelling excavations. Site I Site II Site III 61 26 13 34 10 31 28 55 65 18 63 22 78 15 50 19 26 Shall we reject or not reject the claim that there is no difference in population mean Mimbres classic black-on-white sherd counts for the three sites? Use a 10% level of significance. (a) What is the level of significance

(b) Find SS_TOT, SS_BET, and SS_W and check that SS_TOT = SS_BET + SS_W. (Use 3 decimal places.)

SSTOT	=
SSBET	=
SSW	=

Find d.f._BET, d.f._W, MS_BET, and MS_W. (Use 2 decimal places for MS_BET, and MS_W.)

dfBET	=
dfW	=
MSBET	=
MSW	=

Find the value of the sample F statistic. (Use 2 decimal places.)


What are the degrees of freedom?
(numerator)
(denominator)

(c) Find the P-value of the sample test statistic. (Use 4 decimal places.)

(f) Make a summary table for your ANOVA test.

Source of Variation	Sum of Squares	Degrees of Freedom	MS	F Ratio	P Value	Test Decision
Between groups
Within groups
Total

Answer 1

No of observations seems less and not equal for all three sites,

Could you please check it again.

Following are the calculations i have given by removing last observation(observation 6 from all three sites)

Group 1 61 26 13 34 10 Group 2 Group 3 78 15 50 19 26 28 65 18 63 144 28.8 5822 20.462 1674.8 229 45.8 12327 21.441 1838.8 188 37.6 9846 26.35 2777.2 Sum Average- St. Dev. -

The total sample size is N 15. Therefore, the total degrees of freedom are dftotal -15-1 14 Also, the between-groups degrees of freedom are dfbetween 3-12, and the within-groups degrees of freedom are dfwithin = dftotal-dfbetueen-14 2 = 12 First, we need to compute the total sum of values and the grand mean. The following is obtained Σ x,-144 + 229 + 188-561 Also, the sum of squared values is Σ x, 5822 + 12327 + 9846-27995 Based on the above calculations, the total sum of squares is computed as follows Xij-2 5612 27995 3017013.6 total

The within sum of squares is computed as shown in the calculation below: SSwithin SSuithingroups 1674.8+1838.8+2777.2-6290.8 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 722.8361.4 between between SSuithin 6290.8 MSuithin dfwithin 12 524.233 Now that sum of squares are computed, we can proceed with computing the mean sum of squares , SSbetween 722.8 MSbetweendfbetueen SSuithin 6290.8 MSuithin dfuithin 12524.233

Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows: MSritlsin vithin524.233 0.689 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested 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 0.1, and the degrees of freedom are dfı - 2 and dfe 2, therefore, the rejection region for this F-test is R F F F 2.807 (3) Test Statistics MStetween361.4 MS uithin 524.233 0.689

F = 0.69, (With two Decimal Places)

(4) Decision about the null hypothesis Since it is observed that F-0.689 < F, null hypothesis is not rejected. 2.807, it is then concluded that the Using the P-value approach. The p-value is p 0.5207, and since p 0.1, it is concluded that the null hypothesis is not rejected. 0.5207 > to) Corctars 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 0.1 significance level

Result Details Source df MS Between- treatments 722.8 361.4 F-0.689 Within- treatments 6290.8 12 524.2333 Total 7013.6 14