Answer to Question #186104 in Statistics and Probability for Arijit Das

Question

Two sample drawn from two different populations

Mean Standard deviation Size

Sample- I 165 16 65

Sample- II 150 15 52

Test at 5% level of significance that the difference of mean is 12

Answer 1

Two sample t-test assuming equal variances.

Null hypothesis $H_0:\mu_1=\mu_2$ .

Alternative hypothesis $H_a:\mu\ne\mu_2 .$

Pooled variance: $s^2=\frac{64*16^2+51*15^2}{65+52-2}=171.82.$

Test statistic: $\frac{165-150-12}{\sqrt{171.82(\frac{1}{65}+\frac{1}{53})}}=1.04.$

Degrees of freedom: $df=65+52-2=115.$

P-value: $p<0.3024.$

Since the P-value is greater than 0.05, fail to reject the null hypothesis.

There is no sufficient evidence that the population difference of means is not 12.