Answer to Question #284049 in Statistics and Probability for Bilal

Question

Given the data below test the hypothesis that the mean of three populations are equal.let alpha=0.05

Sample-1 sample-2 sample-3

40 70 45

50 65 38

60 66 60

65 50 42

Answer 1

Solution:

To test the mean of three population we use Analysis of Variance (ANOVA).It is used for testing the equality of three or more population.

μ_{1 -} Mean of population 1

μ_{2 -} Mean of population 2

μ_{3 -} Mean of population 3

Step 1: Formulate the null and alternative hypothesis.

The null Hypothesis (H₀) = μ₁ =μ₂ =μ₃

Alternative Hypothesis(H₁) = at least one of the means is significantly different from the others.

Σ X_A = 215, Σ X_B = 251,   Σ X_C = 185

Σ X_A² = 11925 , Σ X_B² = 15981, Σ X_C² = 8833

Let k be the number of different samples = 3

n = Total number of samples = n₁+n₂+n₃ =4+4+4 = 12

Where n₁-Number of samples in population A

          n₂ -Number of samples in population B

          n₃ - Number of samples in population C

Data table

sum of squares between samples

(∑T_i²/ni)= (215)²/4+(251)²/4 +(185)²/4 = 11556.25 + 15750.25 + 8556.25 = 35862.75

SSB = (∑T_i²/ni) - (∑x)²/n =    35862.75 - 423801/12 = 35862.75 - 35316.75 = 546

sum of squares within samples

SSW = ∑x²-(∑Ti²/ni) = 36739 - 35862.75 = 876.25

Total sum of squares

SST=SSB+SSW

      = 546 + 876.25

SST = 1422.25

variance between samples

MSB=SSB/(k-1)

MSB = 546/(3-1) = 546/2 = 273

variance within samples

MSW = SSW/(n-k)

MSW  = 876.25 /9 = 97.36

test statistic F for one way ANOVA test

F=MSB/MSW

F = 273/97.36

F =2.80

The degree of freedom between samples

k-1=2

The degree of freedom within samples

n-k = 12-3=9

ANOVA table

F(k -1,n-k) = F(2,9) at 0.05 level of significance = 4.256

Calculated F value = 2.80

Table value of F = 4.256

Caculated value < Table value

So, H₀ is accepted, Hence there is no significant difference between mean of three population.