Question

1 -15 points Understa12 19.00 My O ur Teacher N isamuter Scient who is formulating a large and complicated program for a type of data processing. She way of storing and retrieving t o b e so hard drive the programs three level of cance to test the hypothesis that the mean processing times the same for each method Casily the problemas en CN- test of t endence or homogenety, C r e es-of -ware for best ating F o rce One-way ANOVA, Twewy ANOVA, therefore the O reo-oft ve the value of the level of significance State them and MO HP - #2 HP -- POR the : A the means we : Althree mense Ey means are N O Mansare und the same st . (Round your Face to two decimal places d e width P-value of the samples Since ther e that the f act 0.01,

Answer 1

The detailed solution given below. Kindly cross-checks the number. The picture was not clear. You may need to rework it accordingly

Hard Disk : 8.2, 9, 7.5, 8.2 Cloud : 7.8, 8.7, 73, 7.4 Disk: 7.1, 6, 6-4, 6.2

classify the problem
One way anova

level of signficance : 0.01

Null and alternate hypothesis
option 4.
Ho
H1 : Not all the means are equal

test statistic = 10.50

pvalue
option 4
0.001 < P-value < 0.010

iv conclusion
Since the pvalue is less than the level of signficance at alpha - 0.01, we reject Ho

v. at the 1% level of signficance there sufficient evidecen to conclude that the mean times to execute the programs are not all equal.

We are given Hard Disk: X1 = 8.225 S} = 0.3758333 n1 = 4 Cloud: X, = 7.8 S = 0.4066667 na = 4 Disk: X3 = 6.425 S; = 0.2291667 n3 = 4 Step 1 - State the hypotheses Ho: M1 = H2 = 113 H : At least one mean is different form the other. Step 2 Find the critical value No. of group = K = 3 Total no. of observation = N = 4+4+4 = 12 dfN = K -1 = 2 dfD = N-k= 9 a = 0.05 Using a and df = 2, df = 9, we find the critical value Feritical = 0.7494 Step 3 Compute the test value a. Grand mean T . (Xixni)+(XzXn2)+(x,xn) N AGM = 10 (8.225 x 4)+(7.8x4)+(6.425 x 4) - = 7.4833 12 b. Sum of squares within and between group S Sbetween = n1(X1 - AGM) + n2(X, - ŽGM)? +ng(#3 - XGM) SSbetween = 48.225 – 7.4833)² + 4(7.8 – 7.4833)² + 4(6.425 – 7.4833)? = 7.0817 S Swithin = (n1 - 1) S +(n2 – 1)S: +(nz – 1)S SSwithin = (4 – 1)0.3758333+ (4 – 1)0.4066667 +(4 – 1)0.2291667 = 3.035

C. Variance within and between group Sbetween = Seneste = 70817 = 3.5408 S ithin 3.035 Perthan (921-1)+(n2-1)+(13-1) (4-1)+(4-1)+(4-1) 1 = 0.3372 d. F statistics Fstat = Sertaneen = 3.5408 Situs 0.23710.5006 e. Pvalue Using Fstat and df = 2, dfD = 9, we find the pvalue Pvalue = 0.0044 Step 4 Since the pvalue is less than the 0.05, we Reject Step 5 There is enough evidence to reject the null hypothesis and conclude that at least one mean is different from the others. Summary Table Sum of squares d.f. Mean square Pvalue Between 7.0817 10.5006 0.00439999999999996 2 9 3.5408 0.3372 Within 3.035 Total 10.1167