Question

To compare the nicotine content of two quality cigarettes A and B, A examined 60 specimens and B examined 40 specimens.

cigarettes	average	dispertion
A	15.4	3
B	16.8	4

Is there a difference in the nicotine content of the two cigarettes at the 5% significance level?

Answer 1

In the usual notation we are given

n1=60 x1bar=15.4. S1=3

n2=40. X2bar=16.8. S2=4

Null hypothesis: there is no significant difference between the average nicotine content of quality cigarettes I.e mu1=mu2

V/s

Alternative hypothesis:mu1< mu2 (left tailed test)

S.E(x1bar-x2bar)=√(s1^2/n1+s2^2/n2)

=√(9/60+16/40)

√0.55

=0.7416

Test statistic under Ho is

Z=(x1bar-x2bar)/S.E.(x1bar-x2bar)

Z=(15.4-16.8)/0.7416

Z=-1.4/0.7416

Z=-1.8878

Z=-1.88

Critical region for one tailed test the critical value of z at 5℅ level of significance is 1.645

The critical region for left tailed test thus consists of all values of z<=-1.645

Conclusion : calculated value of z(-1.88) is less than critical value (-1.645) it is significant at 5℅ level of significance

Hence null hypothesis is rejected at 5℅ level of significance

We concluded that average nicotine content of quality cigarettes B is higher than those quality cigarette A .