Carry out a test at level = 0.05 to decide whether true average activity differs
ID: 3070767 • Letter: C
Question
Carry out a test at level = 0.05 to decide whether true average activity differs for the three groups. If appropriate, investigate differences amongst the means with a multiple comparisons method. (Use i = 1, 2, and 3 respectively.) State the appropriate hypotheses.
The accompanying summary data on skeletal-muscle CS activity (nmol/min/mg) appeared in an article Old Young Sedentary Active 8 Old Sample size 10 Sample mean 46.88 Sample sd 10 48.11 58.64 8.43 7.16 5.59 Carry out a test at level 0.05 to decide whether true average activity differs for the three groups. If appropriate, investigate differences amongst the means with a multiple comparisons method. (Use-1, 2, and 3 respectively.) State the appropriate hypotheses Ha: all three u's are equal Ha: at least two 's are equal at least two 's are unequal Ha: all three 's are unequal Calculate the test statistic. (Round your answer to two decimal places.) f-7.72Explanation / Answer
I think you need multiple comparison solution , we used here
the fisher least square difference=LSD=sqrt(MSE(1/n1+1/n2))*t(alpha, error df)
w12=sqrt(52.79*(1/10 + 1/8))*t(0.05,25)=3.45*2.06=7.11
w13=sqrt(52.79*(1/8 + 1/10))*t(0.05,25)=3.45*2.06=7.11
w23=sqrt(52.79*(1/10 + 1/10))*t(0.05,25)=3.25*2.06=6.70
(last part) following pairs are significantly different
young and old active
old sedimenty and old active
difference between young and old sedimentry=|(46.88-48.11)|=1.23 <w12 (=7.11)
this pair does not differ significantly
difference between old sedimentry and old active=|(48.11-58.64)|=10.53>w23(=7.11)
this pari differ significantly
difference between young and old active =|(46.88-58.64)|=11.76 > w13(=6.7)
this pair differ significantly
following information has been generated using ms-excel
within SS between SS Group nj mean(xj-) s2 nj*xj- (n-1)s2 (xj--x-) nj(xj--x-)2 1 10 46.88 51.2656 468.8 461.3904 -4.55143 207.15502 2 8 48.11 31.2481 384.88 218.7367 -3.32143 88.255102 3 10 58.64 71.0649 586.4 639.5841 7.208571 519.63502 sum 28 153.63 153.5786 1440.08 1319.7112 -0.66429 815.045143 grand mean(x-) 51.43143 ANOVA SOURCE DF SS MS F CRITICAL F(0.05) p-value BETWEEN 2 815.05 407.5226 7.72 3.39 0.00245 WITHIN(ERROR) 25 1319.71 52.78845 TOTAL 27 2134.76