A scientist working for a large agriculture company is interested in comparing the effect of various...

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A scientist working for a large agriculture company isinterested in comparing the effect of various feed additives on thegrowth of chickens. Chickens were given feed supplemented witheither soy, cornmeal, whey, linseed, or cricket flour. Theircurrent diet is feed with a soy supplement. After 12 weeks on thediet, each chicken was weighed and the value (in grams) wasrecorded in the table below. Analyze the data to determine if thereis a difference in chicken weight between the different additivesand if so, which supplement is the most effective.

Supplement Type

Soy

Cornmeal

Whey

Linseed

Cricket

43.5

15.2

50.4

115.9

17.1

60.1

23.4

63.2

75.5

21.5

45.1

19.7

91.3

113.4

13.5

47.4

13.4

36.3

53.4

13.1

18.1

16.9

87.4

103.5

25.3

29.1

11.4

27.7

136.6

32.3

a. Was a pretest performed? If so, fill in the values in thetable.

Test type

H0

HA

Crit/Calc or exact p-value

If not, explain why:

b. What was the conclusion of your pre-test? Do you need totransform your data? If so, fill in the transformation you used andyour new critical/calculated value or new p-value.

Conclusion:

Transformation

New Calc/Crit or p-value

c. What are the null and alternative hypotheses for your maintest?

d. Complete the ANOVA table:

H0

HA

V ariance source

df

SS

MS

F

P-value

Among

Within

Total

e. What conclusions can you draw? Do you need to do any post-hoctesting?

f. If you need to do post-hoc testing, fill in the blank cellsin the table below with: which post- hoc test you chose and thep-values for each pair of comparisons. Note: the format of thetable is generic and saves space; it is not meant to imply aspecific test.

Post-hoc Test:

Soy

Cornmeal

Whey

Linseed

Cricket

Linseed

Whey

Cornmeal

g. Plot your data. Based on the results of your ANOVA andpost-hoc testing, what is your biological conclusion? Use the plotto be as specific as possible.

Answer & Explanation Solved by verified expert
4.3 Ratings (778 Votes)
Ans a A pre test to check the normality assumption has beenperformed Here we put the values in R software and check fornormality by ShapiroWilk test Here we test the population is normally distributed vs the population is not normalThe R code is given byx1c435601451474181291x2c152234197134169114x3c504632913363874277x4c1159755113453410351366x5c171215135131253323xcx1x2x3x4x5shapirotestxThe results are given byShapiroWilk normality testdata xW 086044 pvalue 0001034b So here pvalue0001034 which is less than the level ofsignificance 005say So we reject the null    See Answer
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