Question

There is an archaeological study area located in southwestern New Mexico. Potsherds are broken pleces of prehistoric Native American clay vessels. One type of painted ceramic vencel in called Mimbres classic black-on-white. At three different sites, the number of such sherds was counted in local dwelling excavations Site I Site 1 site II1 18 30 71 34 20 17 78 58 26 61 15 10 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 1% level of significan a) What is the level of significance? 0.01 State the null and alternate hypotheses Ho: μι " μ2 " μ,; H. All three means are different. Mal μ.-»,-»,; : At least two means are equal, He: μ.-μ: " μί H1 : Not all the means are equal. Mp μ.-», " Poi H. Exactly two means are equal. ( Make a summary table for your ANOVA test. (Round your answers for SS to three decimal places, your MS and F Ratio to two decimal places, and your P-value to four dei places.) Degrees of Source of Variation Sum of SquaresFreedom Test Decision MS P Value Ratio Between groups 050p-value 100 025 p value

Answer 1

Following is the calculation for One-Way ANOVA test:

The following table is obtained Group 1 68 34 20 17 78 58 26 Group 2 Group 3 18 30 71 28 61 15 10 155 38.75 7651 23.415 1644.75 4 301 43 16593 24.664 3650 166 27.667 7074 22.277 2481.3333333333 Sum Average- St. Dev - The total sample size is N 17. Therefore, the total degrees of freedom are dftotal 17-1-16

Also, the between-groups degrees of freedom are dfbetween degrees of freedom are: 3-1 2, and the within-groups dfwithin=dftotal_dfbetween-16 2=14 First, we need to compute the total sum of values and the grand mean. The following is obtained Xj = 301 + 155 + 166 = 622 1.0 Also, the sum of squared values is = 16593 + 7651 + 7074 = 31318 Based on the above calculations, the total sum of squares is computed as follows 6222 total 2.3 1.J The within sum of squares is computed as shown in the calculation below Σ sstrithingroups-3650+ 1644.75+ 2481.3333333333 = 7776.083 SSwithin The between sum of squares is computed directly as shown in the calculation below:

SS_between​ =   SS_total​ - SS_within

SS_between​ =  8560.118 - 7776.083

SS_between​ =  784.034

Now that sum of squares are computed, we can proceed with computing the mean sum of squares 784.034 392.017 MSbetween--etueen MSSSuithin 7776.083 rithin = dfu ithin =-14 00 .435 Now that sum of squares are computed, we can proceed with computing the mean sum of squares S Sbetuwee MS tnen 784.034 392.017 MSSSuithin 7776.083 dfeithin4555.435 Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows -·-betueen 392.017 MSwithin555.435 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.

Based on the information provided, the significance level isa 0.01, and the degrees of freedom are dfi 2 and df) 6.515 (3) Test Statistics 2, therefore, the rejection region for this F-test is R {F : F > F, F MShetreen 392,017 Swithin 555.435 ith 555.4350.706 (4) Decision about the null hypothesis Since it is observed that F-0.706 < F, rejected 6.515, it is then concluded that the null hypothesis is not Using the P-value approach: The p-value is p 0.5105, and since p 0.51050.0, i is concluded that the null hypothesis is not rejected. 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 α-0.01 significance level.

Answer (a)

The level of significance is 0.01

The following null and alternative hypotheses need to be tested:

Ho: μ₁ = μ₂ = μ₃ = μ₄; Ha: Not all means are equal

Answer (b)

SS_TOT = 8560.118

SS_BET = 784.034

SS_W = 7776.083

df_BET = 2

df_W = 14

MS_BET = 392.017

MS_W = 555.435

The value of sample F statistics = 0.706

Degrees of Freedom

df_N = 2

df_D 14

Summary Table for ANOVA test

Source	Sum of Squares	Degrees of Freedom (df)	Mean Square	F Ratio	P-value	Test Decision
Between	784.034	2	392.017	0.706	p-value > 0.100	Fail to reject null hypothesis
Within (error)	7776.083	14	555.435
Total	8560.118	16

There is not enough evidence to reject the claim that there is no difference in population mean for 3 sites, at α=0.01 significance level.