Question

Assume that both populations are normally distributed. ​a) Test whether mu 1 not equals mu 2 at the alpha equals 0.05 level of significance for the given sample data. ​b) Construct a 95​% confidence interval about mu 1 minus mu 2. Table: Sample 1 Sample 2:

n= 16 16

x bar= 12.3 14.1

s= 3.5 3.5

Answer 1

a)

null hypothesis Alternate Hypothesis 0 Hi-H2 0 degree of freedom (n,+n2-2) V- 30.000 for 0.05 level with two tailed test and n-1-30 degree of freedom, critical value of t- Decision rule 2.042 reject Ho if absolute value of test statistic t>2.042 first sample second sample 12.300 3.500 16.000 14.1000 3.5000 16.0000 X2 S1 52 pool. vars ((n-1)s 1+(n2-1)*s 2/(n,tnz-2)- 12.2500 Point estimate 1.8000 std error of difference-se- Sp*d(1/n1+1/n2) 1.237 test stat t =

Decision: as test statistic is not in rejection region we fail to reject null hypothesis

Conclusion:we do not have sufficient evidence to conclude that there is a  significant difference between population mean

b)

for 95 % CI & 30 df value of t=				=	2.0420
margin of error E=t*std error                            =					2.5268
lower confidence bound=mean difference-margin of error =					-4.327
Upper confidence bound=mean differnce +margin of error=					0.727