A dairy scientist is testing a new feed additive. She chooses 13 cows at random
ID: 3150690 • Letter: A
Question
A dairy scientist is testing a new feed additive. She chooses 13 cows at random from a large population of cows. She randomly assigns nold = 8 to get the old diet, and nnew = 5 to get the new diet including the additive. The cows are housed in 13 separated pens and each gets separate feed, with or without additive as appropriate. After two weeks, she picks a day and milks each cow using standard procedures and records the milk produced in pounds. The data are below: Old Diet: 43, 51, 44, 47, 38, 46, 40, 35 New Diet: 47, 75, 85, 100, 58 Let new and old be the population mean milk productions for the new and old diets, respectively. She wishes to test: H0 :new old =0 vs. HA :new old =0, using = 0.05.
(a) Are the two populations paired or independent? Explain your answer.
(b) Graph the data as you see fit. Why did you choose the graph(s) that you did and what does it (do they) tell you?
(c) Choose a test appropriate for the hypotheses above, and justify your choice based on your answers to parts (a) and (b). Then perform the test by computing a p-value, and making a reject or not reject decision.
(d) Finally, state your conclusion in the context of the problem.
Explanation / Answer
a.
Independent
c. t Test of Difference Means for Unequal Variance
Null Hypothesis , There Is No-Significance between them Ho: u1 = u2
Alternate Hypothesis, There Is Significance between them - H1: u1 != u2
Test Statistic
X(Mean)=43
Standard Deviation(s.d1)=5.1824 ; Number(n1)=8
Y(Mean)=73
Standard Deviation(s.d2)=21.0832; Number(n2)=5
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =43-73/Sqrt((26.85727/8)+(444.50132/5))
to =-3.123
| to | =3.123
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 4 d.f is 2.776
We got |to| = 3.12335 & | t | = 2.776
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -3.1233 ) = 0.035
Hence Value of P0.05 > 0.035,Here we Reject Ho
d.
Reject Ho
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