Question

s. (25) A 2012 study examined the effect of herbal treatments on lactation in buffalo cows. 30 cows were assigned to three different treatments. Group A received echinacea before calving and a m chelidonium Group C wa and lycopodium s the control at 7 day intervals after calving. Group B received only the post-calving treatment, and group. Several variables were measured, but the table below shows only the estimated data ilk yield day in units of kilograms Group A Groups Group 12? 11.9: 11.3 4.3 1.7| 10.5 89 114 160 D 10.5 95 63 105 15010 109 93 102 b.Compute and s a. Compute the summary statistics: Mean Std. Dev Group A Group B Group C d. Compute the P-value. c. Compute the test statistic (F) tement and the interpretation in context Draw the appropriate conclusions (both the mathematical sta e. 2.

Answer 1

a.

The mean and standard deviation of data is,

	Mean	Std.Deviation
Group A	10.14	2.88
Group B	9	3.73
Group C	9.75	1.40

(b)

S_p² = $\sum$ (ni-1) S²_i   = (10 -1) * 2.88² + (10 -1) * 3.73² + (10 -1) * 1.4² = 217.5057

Grand mean, $\bar{\bar{X}}$ = (10.14 + 9 + 9.75)/3 = 9.63

S²_treatment = $\sum$ ni * (Mean_i - $\bar{\bar{X}}$ )²

= 10 * (10.14 - 9.63)² +  10 * (9 - 9.63)² +  10 * (9.75 - 9.63)² = 6.714

S²_total = S_treatment + S_p = 6.714 + 217.5057 = 224.2197

c.

Degree of freedom of treatment, dT = Number of groups - 1 = 3 - 1 = 2

Degree of freedom of error, dE = Number of obsrvations - Number of groups = 30 - 3 = 27

MSTR = S²_treatment / dT = 6.714 / 2 = 3.357

MSE = S_p² / dE = 217.5057 / 27 = 8.056

F = MSTR / MSE = 3.357 / 8.056 = 0.4167

d.

Degree of freedom for F statistic is 2, 27

P-value = P[F > 0.4167] = 0.6634

e.

As, p-value is greater than significance level od 0.05, we fail to reject the null hypothesis and conclude that there is no significant evidence that atleast one of the means of a group is different from each other and there is no significant effect of herbal treatment on lactation in buffalo cows.