✔ 最佳答案
sample mean x_bar1=74, standard deviation 1 =8, n1=40
sample mean x_bar2=78, standard deviation 2 =7, n2=50
null hypothesis: E(x1)-E(x2)=0
alternative: E(x1)-E(x2) not equal to 0
Since standard deviations are known, compute Z
Z=[(x_bar1-x_bar2)-(E(x1)-E(x2))]/ sqrt{[(standard deviation 1)^2/n1]+[(standard deviation 2)^2/n2]}
Z=[74-78-0]/ sqrt[(64/40) + (49/50)]
= -4/ sqrt(2.58)
= -2.49
(i) significant level=5% (2-sided or 2-tail)
critical value= +/-1.96
|Z| = 2.49 > 1.96
reject null hypothesis.
so there is significant difference between the performance of the two classes if level of significance is 5%.
(ii) significant level=1% (2-sided or 2-tail)
critical value= +/- 2.575
|Z| = 2.49 < 2.575
not reject null hypothesis, that is, (E(x1)-E(x2))=0
so no significant difference between the performance of the two classes if level of significance is 1%.