Question

a.Calculate the total sum of squares​ (SST) and partition the SST into its two​ components, the sum of squares between​ (SSB) and the sum of squares within​ (SSW).

b. Use these values to construct a​ one-way ANOVA table.

c. Using alpha equals0.05, what conclusions can be made concerning the population​ means?

Sample_1   Sample_2   Sample_3
3                       1              7
2                       3              6
16                     5              3
                                           8

Determine the values.

SSTequals

SSBequals

SSWequals

​b) Complete the​ one-way ANOVA table below.

Source	Sum of Squares	Degrees of Freedom	Mean Sum of Squares	F
Between	nothing

Within

nothing

Total	nothing

nothing

​(Type integers or decimals. Round to three decimal places as​ needed.)

​c) Let mu 1 mu 2 and mu 3 be the population means of samples​ 1, 2, and​ 3, respectively. What are the correct hypotheses for a​ one-way ANOVA​ test?

What is the critical​ F-score,

What is the correct conclusion about the population​ means?

Answer 1

​c) Let mu 1 mu 2 and mu 3 be the population means of samples​ 1, 2, and​ 3, respectively. What are the correct hypotheses for a​ one-way ANOVA​ test?

H₁: At least one of the mean is different from the others

	x1	x2	x3
	3	1	7
	2	3	6
	16	5	3
			8	Total
sum	21	9	24	54
sum of square	269	35	158	462

Grand total = =54

N=10

T1, T2 , T3 are group totals

SSE= SST-SSB   = 170.4-26.4 =144.0

ANOVA table
Source	SS	df	MS	F
Between	26.40	2	13.200	0.642
within	144.00	7	20.571
Total	170.40	9

What is the critical​ F-score,

Table value of F with (DF1=2, DF2=7 ) =4.737

Calculated F=0.642 < 4.737 the table value.

The null hypothesis is not rejected.

What is the correct conclusion about the population​ means?

The data indicate there is no significant difference among the three groups