Answer:
Since the calculated value of F = 1.4397 is less than the critical value of
F (9,9)= 2.4403 we conclude that the first instructor's variance is smaller and reject H0.
Step-by-step explanation:
1)Formulate the hypothesis that first variance is equal or greater than the second variance
H0: σ₁²≥σ₂² against the claim that the first instructor's variance is smaller
Ha: σ₁²< σ₂²
2) Test Statistic F= s₂²/s₁²
F= 84.8/ 58.9= 1.4397
3)Degrees of Freedom = n1-1= 10-1= 9 and n2 = 10-1= 9
4)Critical value at 10 % significance level= F(9,9)= 2.4403
5)Since the calculated value of F = 1.4397 is less than the critical value of
F (9,9)= 2.4403 we conclude that the first instructor's variance is smaller and reject H0.
