Question

Summary is obtained from two independent Normal samples

a). Test whether one can assume equal variances.
b) With a suitable test procedure, test for equality of means
.

Sample1 20 28 8 Sample2 40 Sample size Sample mean Sample standard deviation 10

Answer 1

Ans:

a)

Test statistic:

F=8^2/10^2=0.64

df1=20-1=19

df2=40-1=39

p-value=1-FDIST(0.64,19,39)=0.1490

As,p-value is greater than 0.05,we fail to reject the null hypothesis.

There is not sufficient evidence to conclude that there is significant difference in variances.

So,we can assume equal variances.

b)

pooled standard deviation=SQRT(((20-1)*8^2+(40-1)*10^2)/(20+40-2))=9.392

standard error for difference=9.392*sqrt((1/20)+(1/40))=2.572

Test statistic:

t=(28-30)/2.572

t=-0.7776

df=20+40-2=58

p-value=tdist(0.7776,58,2)=0.4400

As,p-value>0.05,we fail to reject the mull hypothesis.

There is not sufficient evidence to conclude that there is significant difference in means.