Question

Preliminary data analyses indicate that it is reasonable to consider the assumptions for one-way ANOVA satisfied. Perform the required hypothesis test using the critical-value approach. At the 0.01 significance level, do the data provide sufficient evidence to support the claim that a difference exists between the population means of the three different brands? The sample data are given below.

Answer 1

Solution : Here we have 3 groups Brand A Brand B and Brand a 5 observatione in Brand A 7 Observations in Brand B 5 observations in Brande N2=7 13= 5 The number of groups K = 3 Total observations N = mit n2 +n3 N = 5+ 7 +5 N=17 ni=5, To Perform ANOVA Define Hypothesis Ho: Mi= Mi = M2 = 43 M2 = M3 There is not significant difference between the population means of the three different Brande Hei There is Synificance difference between between the population means of the three different Brands. step 1: correction factor (CF) = 2 (Ex) N where EX= Sxi + EX2 +EX3

Brand A ·X1 44 47 44 yo 39 Brand B X2 30 32 34 36 38 40 42 X2= 252 Brand C X3 28 27 31 32 36 EXI = 214 X3 = 154 Xi 214 214 – 4208 T2= 252-36 7 X3= 154= 30$ 5 X3 = 30.8 = 42.8 X2 = 36 EX = 214 +252 + 154 = 620 CE = (620) = 22611.765 384400 17 17 CF-2261. 765 Step 2: Total Sum of squeses SST: Exi? - Cof SST = (EX? + EX2? + EX32) – GoF Now we calculate square value of each Observation

X2 1936 2209 1936 1600 1521 X2 900 1024 1156 1296 1444 1600 1764 EX22=9184 X3? 784 729 961 1024 1296 8X²=9202 EX32=4794 SSTE SST= ( Exi?+ EXz?+ 3x3?) - CE 9202 +9184 + 4794) – 22611•765 23180 - 22611.765 568.235 ISST = Step3: Between sum of squares SSB = | Ex1)+ (x2)* +(3X3)? - CE DA D2 n3 SSB = [214) + (252) +(154)* (154)*) –226 11.765 5 7 22611.765 45796 5 63504 7 + 23716 5 = [9159.2 + 9072 + 4743.2] – 22611.765 22974.4 22611.765 = 362.635

SSB = 362. 635 Step 4 Within Sum of Squares (Errors) SSE = SS T- SSB 568.235 - 362.635 = 205.6 SSE = 205.6 Now we know know k= 3 N=17 degkee of freedom (SSB) = K -1 = 3-1 = 2 degull of freedom (SSE) = N-K= 17-3 = 14 dege of freedom (SST) = N-1 = 17 -1 = 16 MSB- 181.318 SSB K-1 3624635 2 MSE- SSE N-K 205.6 14 =_14.686 MSB MSE 1816 318 14.686 =122346 F = 12.346

ANOVA Table Idf MS E Source SS Bet weergroups | 362.635 2 1 B 16 318 12 346 within (essors) 205.6 14 14.686 Total 568.235 16 Calculated F= 12.346 Significance level a = 0.01 Critical value of F at d=0001 for (2,14) degree of freedom is Fover (2,14) = 6:52 [ using f-tables ] Since calculated F > critical F 12.346 6.52 > So, we reject Ho at oool level of signifirane Conclusion: At the oiol significance level, the data provide sufficient evidence to support the claim that a difference exist between the population meons of the three different brands.