Question

There 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
64	26	15
35	17	32
22	54	65
11	68	28
77		16
54		13
20

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 significance.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: μ₁ = μ₂ = μ₃; H₁: Exactly two means are equal.H₀: μ₁ = μ₂ = μ₃; H₁: Not all the means are equal.     H₀:μ₁ = μ₂ = μ₃; H₁: At least two means are equal.H₀: μ₁ = μ₂ = μ₃; H₁: All three means are different.

(b) Find SS_TOT, SS_BET, and SS_W and check that SS_TOT = SS_BET + SS_W. (Round your answers to three decimal places.)

SSTOT	=
SSBET	=
SSW	=

Find d.f._BET, d.f._W, MS_BET, and MS_W. (Round your answers for MS_BET, and MS_W to two decimal places.)

d.f.BET	=
d.f.W	=
MSBET	=
MSW	=

Find the value of the sample F statistic. (Round your answer to two decimal places.)


What are the degrees of freedom?

d.f.N	=
d.f.D	=

(c) Find the P-value of the sample test statistic.

P-value > 0.1000.050 < P-value < 0.100     0.025 < P-value < 0.0500.010 < P-value < 0.0250.001 < P-value < 0.010P-value < 0.001

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

Since the P-value is greater than the level of significance at α = 0.01, we do not reject H₀.Since the P-value is less than or equal to the level of significance at α = 0.01, we reject H₀.     Since the P-value is greater than the level of significance at α = 0.01, we reject H₀.Since the P-value is less than or equal to the level of significance at α = 0.01, we do not reject H₀.

(e) Interpret your conclusion in the context of the application.

At the 1% level of significance there is insufficient evidence to conclude that the means are not all equal.At the 1% level of significance there is sufficient evidence to conclude that the means are all equal.     At the 1% level of significance there is insufficient evidence to conclude that the means are all equal.At the 1% level of significance there is sufficient evidence to conclude that the means are not all equal.

(f) 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 decimal places.)

Source of Variation	Sum of Squares	Degrees of Freedom	MS	F Ratio	P-Value	Test Decision
Between groups					---Select--- p-value > 0.100 0.050 < p-value < 0.100 0.025 < p-value < 0.050 0.010 < p-value < 0.025 0.001 < p-value < 0.010 p-value < 0.001	---Select--- Reject H0. Do not reject H0.
Within groups

Answer 1

Solution:-)

RR Console > x= read.table (file.choosse (), sep=",", header-T) stack x= stack (x) > anova-aov (values ind , data- stack x) summary (anova) Df Sum Sq Mean Sq F value Pr (>F) ind 2 452 226.2 0.402 0.678 Residuals 11 6185 562.3 4 observations deleted due to missingness R Untitled - R Editor x read.table (file.choose ),sep",", header-T) stack_x stack (x) anova-aov (values ind , data= stack x) summary (anova)||

A) The level of significance is 0.01.

B) The null and alternative hypotheses are

\

Ho: 1 2 = 3

That is all the means are equal .

Ha 2143

At least none of them are equal.

C) From above the output of R, we have

SSTOT	6637
SSBET	452
SSW	6185

We have the following results,

d.f._BET = 2 , d.f._W = 11 , MS_BET =   226.2  , MS_W= 562.3

D) The value of the sample F statistic is

226.2 0.402 F 562.3

e) The P-value of the sample test statistic is 0.678 > 0.01

f) As the p-value is greater than significance level, therefore we fail to reject the null hypothesis at significance level

= 0.01.

g) .At the 1% level of significance there is sufficient evidence to conclude that the means are all equal. Therefore null hypothesis is accepted. All the means are equal.

h) The summary table will be

Source of Variation	Sum of Squares	Degrees of Freedom	MS	P-Value	Test Decision
Between groups	452	2	226.2	0.678	Therefore,Do not reject H0.
Within groups	6185	11	562.3	As p-value> 0.01