Activity 7.docx

Activity 7: One way Analysis of variance

Exercise 1 1) H0 :µ1= µ2= 2)

µ3= µ4 HA: µ1≠µ2≠µ3≠µ4

3) 𝑘

𝑛𝑆

1 𝑥̄̄ = ∑ ∑ 𝑥𝑖𝑗 𝑛 𝑖=1 𝑗=1

= (3.43+6.43+5.29+4.29)/4 = 4.86

=(7(3.43-4.86)2+7(6.43-4.86) 2+7(5.29-4.86) 2+7(4.29-4.86) 2)/3 =11.7124

=(6(0.982)+6(1.132)+6(1.112)+ 6(1.112))/24 =1.1754 𝐹 =

𝑀𝑆𝐵 𝑀𝑆𝑊

= (11.7124/1.1754) = 9.964

From excel

Anova: Single Factor SUMMARY Groups Column 1 Column 2 Column 3 Column 4

Count

Sum 7 7 7 7

ANOVA Source of Variation Between Groups Within Groups

SS 35.14286 28.28571

Total

63.42857

24 45 37 30

df

Average 3.428571 6.428571 5.285714 4.285714

Variance 0.952381 1.285714 1.238095 1.238095

MS F P-value F crit 3 11.71429 9.939394 0.000191 3.00878657 24 1.178571 27

4) P value 0.000191 5) The p value is less than 0.05 so the null hypothesis is rejected at 5% level of significance. 6) There is sufficient evidence to conclude that the means from the different regions are not the same.