Question

An experiment was conducted to see if there is any difference in the amount of weight gained by two groups of pigs fed + Feed # Pigs Mean Weight Gain Standard Deviation Diet A 36 55 3 Diet B 49 53 5.6 1. Construct a 95% confidence interval for the difference of means weight gain using 76 as the degree of freedom. 2. Assume that the variance of the 2 populations are equal, test the hypothesis that pigs fed with Diet A gain more weight than with Diet B at the 0.05 level. 2.

Answer 1

Answer:

1. The confidence interval for the difference of means weight gain using 76 as the degree of freedom is -1.664, 5.664

2.The calculated value of t does not lie in the critical region for both d.f 76 and 83 . Therefore we accept our null hypothesis that pigs fed with Diet A gain more weight than with diet B at the 0.05 level

Explanation:

# Pigs Mean Weight Gain Standard Deviation

Diet A 36 55 3

Diet B 49 53 5.6

(x`1- x` The 2) ± t∝/2 (v) . Sp √1/n1 + 1/n2

where v= d.f= 76

Here the difference between the sample means is x`1- x`2= 55-53= 2

The pooled estimate for the common variance σ² is

Sp² = 1/n1+n2 -2 [ ∑ (x1i - x1`)² + ∑ (x2j - x`2)²]

= 1/36 +49-2 [ 55²+53²]

Sp = √3025 +2809/83 = 8.38

t (0.025) (76) = 1.992

(x`1- x`2)± t∝/2 (v) . Sp √1/n1 + 1/n2

(2 )± 1.992 (8.38) √ 1/36 + 1/49

(2 )± 3.664

-1.664, 5.664

Par t 2:

The null and alternate hypotheses are

H0 : u1 > u2 against Ha: u1 ≤ u 2

Significance level alpha = 0.05

The critical region t ≥ t (0.05, 76) = 1.665

D.f = n1+n2- 2= 83

The critical region t ≥ t (0.05, 83) = 1.663

t = (x`1- x` ) /. Sp √1/n1 + 1/n2

t= 2/ 8.38 √1/36+ 1/49

t= 2/1.839

t= 1.087

The calculated value of t does not lie in the critical region for both d.f 76 and 83 . Therefore we accept our null hypothesis that pigs fed with Diet A gain more weight than with diet B at the 0.05 level