Question

2. Researchers are investigating the affects diet can have on the brain mass of mice. Two samples of mice are fed diets A and B and after a set period of time the weight of their brains (in mg) is measured. Diet A Diet B I = 537 72 = 685 $1 = 266 s2 = 385 ni = 116 n2 = 52 (a) At the 1% significance level, do the data provide sufficient evidence to conclude that diet A is less nutritious (leads to less brain mass)? Do not assume that the standard deviations of the populations are equal. (Use the P-value approach).

Answer 1

To Test :-

H0 :- µ1 = µ2
H1 :- µ1 < µ2

Test Statistic :-
$t = (\bar{X_{1}} - \bar{X_{2}}) / \sqrt{ ( S_{1}^{2} / n1) + (S_{2}^{2} / n2)}$
$t = ( 537 - 685 ) / \sqrt{ ( 266^{2} / 116) + ( 385^{2} / 52)}$
t = -2.5159

Decision based on P value
P - value = P ( t > 2.5159 ) = 0.007
Reject null hypothesis if P value < α level of significance
P - value = 0.007 < 0.01 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

There is sufficient evidence to support the claim that diet A is less nutritious than diet B.

Part b)

Point estimate = $\( \bar{X}_{1} - \bar{X}_{2}) = -148.00$

Confidence interval :-

(X1 – X2) + ta/2,DF) (S}/n1) + (S}/n2)

DF = 73

t(α/2, DF) = t(0.02 /2, 73 ) = 2.379
$( 537 - 685 ) \pm t_{ 0.02/2 , 73 }\sqrt{ ( 266^{2}/ 116 ) + ( 385^{2}/ 52 )}$
Lower Limit = $( 537 - 685 ) - t_{ 0.02/2 , 73 }\sqrt{ ( 266^{2}/ 116 ) + ( 385^{2}/ 52 )}$
Lower Limit = -287.946
Upper Limit = $( 537 - 685 ) + t_{ 0.02/2 , 73 }\sqrt{ ( 266^{2}/ 116 ) + ( 385^{2}/ 52 )}$
Upper Limit = -8.054
98% Confidence interval is ( -287.946 , -8.054 )

Margin of Error = $t_{\alpha /2 , DF}\sqrt{ (S_{1}^{2}/n1) + (S_{2}^{2}/n2)} = 139.946$