Question

Twenty laboratory mice were randomly divided into two groups of 10. Each group was fed according to a prescribed diet. At the end of 3 weeks, the weight gained by each animal was recorded. Do the data in the following table justify the conclusion that the mean weight gained on diet B was greater than the mean weight gained on diet A, at the α = 0.05 level of significance? Assume normality. (Use Diet B - Diet A.) Diet A 5 8 10 6 6 10 13 8 5 9 Diet B 9 10 7 23 12 15 13 7 17 12 (a) Find t. (Give your answer correct to two decimal places.) (ii) Find the p-value

Answer 1

ANSWER:

Given that,

Diet A (x)	5	8	10	6	6	10	13	8	5	9
Diet B(y)	9	10	7	23	12	15	13	7	17	12

= x / = (5+8+10+6+6+10+13+8+5+9) / 10 = 80/10 = 8

= sqrt ( (x-)^2 / (-1))

= sqrt (((5-8)^2+(8-8)^2+(10-8)^2+(6-8)^2+(6-8)^2+(10-8)^2+(13-8)^2+(8-8)^2+(5-8)^2+(9-8)^2)/(10-1))

= 2.5819

= y / = (9+10+7+23+12+15+13+7+17+12) / 10 = 125/10 = 12.5

= sqrt ( (y-)^2 / (-1))

= sqrt (((9-12.5)^2+(10-12.5)^2+(7-12.5)^2+(23-12.5)^2+(12-12.5)^2+(15-12.5)^2+(13-12.5)^2+(7-12.5)^2+(17-12.5)^2+(12-12.5)^2)/(10-1))

= 4.9046

(a) Find t

t = (-) / sqrt(/ + /)

t = (8-12.5) / sqrt((2.5819)^2/ 10+(4.9046)^2/10)​​​

t = -4.5 / 1.7527495191840733

t = - 2.567394799

t = - 2.57

(ii) Find the p-value:

Degree of freedom = df = +-2 = 10+10-2 = 18

p-value = 0.01927

approx

p-value = 0.0193

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