Question

	Sample 1	Sample 2
n (size)	15	15
average	1345	1210
stdev	31	28
variance	961	784
Pooled variance	872,50
st.err	10,786
t	12,516
alpha	0,05
tα	1,701
p-value	0,0000
Confidence intervals
lower	116,6519537
upper	153,3480463

An automobile company is considering the purchase of batteries in bulk from two suppliers. A sample of 15 batteries is randomly selected from each supplier and tested .

The mean lifetimes are:

?1=1345 h????   ?1=31 h????   ?1 =15

?2=1210 h????   ?2=28 h????   ?2 =15

a) State the null and alternative hypotheses

b) What should the automobile company conclude if the test is conducted at the 0,05 significance level?

Answer 1

Given,

?1 bar =1345 h????   ?1=31 h????   ?1 =15

?2 bar =1210 h????   ?2=28 h????   ?2 =15

F test for equal variance

F = S1^2/S2^2 = 1.2258

P value = 0.3543 (Use F table)

P value > 0.05, Do not reject null hypothesis of equal variances

Two sample t test:

a) Hypothesis :		α=	0.05
		df	28	n1+n2-2
Ho:	μ1 = μ2
Ha:	μ1 not = μ2
t Critical Value :
tc	2.048407142	T.INV.2T(D1,D2)	TWO
ts	< for -	tc	TWO	To reject
ts	> for +	tc	TWO	To reject
Test :
Sp^2	872.5	((n1-1)S1^2+(n2-1)S2^2)/(n1+n2-2)
t	12.5164648	(X1 bar-X2 bar )/SQRT(Sp^2*(1/n1 + 1/n2))	Equal vriance
P value :
P	5.47733E-13	T.DIST.2T(ts,df)	TWO
Decision :
P value	<	α	Reject H0

b) Conclusion:

There is enough evidence to conclude that the mean lifetimes of 15 batteries from two suppliers is different at 5% significance level