Question

Here 5.0000080853276E-7 after Background: The butterfat content of random samples of Guernsey, Holstein and Jersey cows are provided below. Each data value represents the percentage butterfat.

Source: Sokal, R. R. and Rohlf, F. J. (1994) Biometry. W. H. Freeman, New York, third edition.

Directions: Perform an analysis of variance (ANOVA) to determine if the differences in the percentage butterfat of the different breeds of cows is statistically significant.

1. Click on the Data button below to display the data. Copy the data into a statistical software package and click the Data button a second time to hide it.

   Data

   Guernsey	Jersey	Holstein
   4.64	4.75	3.22
   5.04	6.54	3.58
   5.68	5.42	3.81
   4.85	4.5	4.17
   4.57	5.25	4.41
   4.63	5.55	3.56
   4.66	5.33	3.34
   3.95	4.76	3.72
   5.05	5.27	3.55
   4.56	5.03	3.65
   5.18	5.61	3.84
   4.52	5.13	3.47
   4.64	5.01	3.44
   4.99	4.85	3.49
   4.7	5.21	3.68
   4.5	4.32	4.18
   5	5.88
   5.27	4.84
   5.4
   6.13

2. Choose the correct null and alternative hypotheses.
   • H0: μ1≠μ2≠μ3H0: μ1≠μ2≠μ3
     Ha: μ1=μ2=μ3Ha: μ1=μ2=μ3
   • Ha: μ1,μ2,μ3Ha: μ1,μ2,μ3 are not all equal.
     H0: μ1=μ2=μ3H0: μ1=μ2=μ3
   • H0: μ1=μ2=μ3H0: μ1=μ2=μ3
     Ha: μ1,μ2,μ3Ha: μ1,μ2,μ3 are not all equal.
   • H0: μ1=μ2=μ3H0: μ1=μ2=μ3
     Ha: μ1≠μ2≠μ3Ha: μ1≠μ2≠μ3
3. Compute the test statistic.
   
   Compute the treatment sum of squares (SSTr) and the error sum of squares (SSE) and use them to complete the following ANOVA table. (Round your answers to 4 decimal places).
   

   Source	S.S.	df	M.S.	F
   Treatment
   Error
   Total

4. Compute the p-value. (Round your answer to 4 decimal places.)
   
   pp-value =
   
5. Interpret the results of the significance test.
   • The p-value provides little evidence against the null hypothesis. The differences in the butterfat content of Guernsey, Holstein and Jersey cows is not statistically significant..
   • The p-value provides strong evidence against the null hypothesis. The differences in the butterfat content of Guernsey, Holstein and Jersey cows is statistically significant.

Answer 1

Answer: ANOVA is used in testing of difference in more than two group means. For the given data of three groups (Guernsey, Jersey & Holstein), the hypothesis set up will be We test Ho: M, = M2 = Mg, ie. there is no significant difference in % of butterfat of the three groups of cows vs Hi: at least two Mis are not equal, i.e. 41,42,43 are not all equal. Now the test statistic is given by E = MST which follows F-distribution under Ho with (Kol, nak) degrees of freedom E MSE due to enor where Kin no of groups & n is total observation. Here K=3 & n= 54. So F F (2,50 under Ho And MST= mean sum of square due to treetments MSE = MST = SST I msESSE where SST = 5 I (81-82 ; I1, 2, 3 whene Xi is mean of observation of ith group & 7 it total sampe mean of all n observations defecerror) i

& mehele SSE Hence I (Xi; – .)? ; whene ni is no. of ous in ith group for the given data, we compute - 20 Fi. = I xis for group 1 (Guernsey) : X= 4.64. - 4.64+, +6:13 - 4.898 For group 2 ( Jersey): 72 = 4.75+ .. +4.645.181 For group 3 ( Holstein): 82= 3.224.. 44.18 - 59,411.3.694 And à = [ Xij - 250.32 4.636 n. 54 Hence SET- (4.898-4.636)²+(5.181-4, 6362 3.694 = (0.26272760.545) + (8 01942) 2 010686 + 0.2970 +0.0074 = 1.2530 I nix,? - ny? SST = 20 X x (4.89812+ 18% (5.181)2+16xQ.694,2 a 54 x (4. 6362 E 479.80 01 +4 83,0868 +218. 3745- 1160.372 = 1181. 269- 20.89713 9 20.0971 And MST = SST = 20.8971 = 10.4486 Kri Similarly SSE = [[ (xi) - Ži)2 (14.64-4,896)2 + ... +613- 4.898)2+ (4.75-5.181)2 t- (4.84-5.181)2 + (3.22- 3.694) 7..+(4.18–3.6.94127 = (-0.25647..+(1.232)2 + (-0.43056 .. +(-0.34056) 22 (-0.47437) ? + +0.48 562572]

(0.066564) + ... + ... 71.517824)+0.18 578+.. +0.115978) + (0.22 5032 + .. a 0.23 5832) = 4.30 632 + 4, 597094 +1.57 0994 SSE = 10.55441 = MSESSE - 10.5544/ .0.2069 so test statistic in F = BST = 10.4486 50.4805] h-k $ 0.2069 Hence A NOVA Table in source S.S. dif. M.S. F Troub 20.8977 2 10.4481 50.4885 (Treatment) Error 10.5544 51 0.2069 Total 31.4515 53 The p-value is given by P-value = Prof F> Feal] = De [ F> 50.4885] = 8.085 x 1013 (using F table) = 0.000 Since p-value < 0.05 (level of significanus, hence we ernect the hull hypothesis Ho. In other words, value provides strong evidence against u The differences in the butterfat insey, Holstein and Jersey cows the null - hypothesis. The differences in of Gernsey, Holstein content is sat statistically significant.