Question

In the book Analysis of Longitudinal Data, 2nd ed., (2002, Oxford University Press), by Diggle, Heagerty, Liang,and Zeger, the authors analyzed the effects of three diets on the protein content of cow’s milk. The data shown here were collected after one week and include 25 cows on the barley diet and 27 cows each on the other two diets:

diet	Protein content of cow's milk.
Barley	3.63	3.24	3.98	3.66	4.34	4.36	4.17	4.4	3.4	3.75	4.2	4.02	4.02	3.9	3.81	3.62	3.66	4.44	4.23	3.82	3.53	4.47	3.93	3.27	3.3
Barley+Lupins	3.38	3.8	3.8	4.59	4.07	4.32	3.56	3.67	4.15	3.51	4.2	4.12	3.52	4.08	4.02	3.18	4.11	3.27	3.27	3.97	3.31	4.12	3.92	3.78	4	4.37	3.79
Lupins	3.69	4.2	4.2	3.13	3.73	4.32	3.04	3.84	3.98	4.18	4.2	4.1	3.25	3.34	3.5	4.13	3.21	3.9	3.5	4.1	2.69	4.3	4.06	3.88	4	3.67	4.27

(a) What is the value of LSD for Barley+Lupins diet and Lupins diet? Use α=0.05.
Round your answer to three decimal places (e.g. 98.765).

(c) What is the absolute value of difference between mean protein content after Barley+Lupins diet and Lupins diet?
Round your answer to three decimal places (e.g. 98.765).

(d) Estimate the standard error for comparing the means using the graphical method. Use minimum sample size.
Round your answer to three decimal places (e.g. 98.765).

Answer 1

Null and Alternative Hypothesis:

H₀: µ_Barley = µ_{Barley+Lupins} = µ_Lupins

H₁: Not all Means are equal

Alpha = 0.05

N = 79

Degress of Freedom:

df_Between = a – 1 = 3-1 =2

df_Within = N-a = 79-3 = 76

df_Total = N-1 = 79-1 = 78

Critical Values:

Time (df_Between, df_Within) : (2,76) = 3.12

Decision Rule:

If F is greater than 3.12, reject the null hypothesis

Test Statistics:

SS_Betwen = ∑(∑a_i)²/n - T²/N = 0.11

SS_Within = ∑(Y)² - ∑(∑a_i)²/n = 12.23

SS_Total = SS_{Betwen ­­}+ SS_Within = 12.34

MS = SS/df

F = MS_effect / MS_error

Hence,

F = 0.06/0.16 = 0.35

Result:

Our F = 0.35, we fail to reject the null hypothesis

Conclusion:

All means are equal.

a)

Post- Hoc Test:

Where:
t = critical value from the t-distribution table
MSW = mean square within

t_critical = 1.99 when at df = 76 alpha = 0.05

MSW = 0.16

n_A = 25

n_B = 27

Hence, LSD = 0.222

c)

Mean _Barley = 3.89

Mean _{Barley+Lupins} = 3.85

Hence, Mod( Mean _{Barley ­­­­­­ ­­­­­}- Mean _{Barley+Lupins} ) = mod(3.89-3.85) = 0.04

Also , this value is less than LSD, hence they are same