Two accounting professors decided to compare the variance of their grading procedures. To accomplish this, they each graded the same 10 exams, with the following results: Mean Grade Standard Deviation Professor 1 79.3 22.4 Professor 2 82.1 12.0 At the 10% level of significance, what is the decision regarding the null hypothesis? A) Reject the null hypothesis and conclude the variances are different. B) Fail to reject the null hypothesis and conclude no significant difference in the variances. C) Reject the null hypothesis and conclude the variances are the same. D) Fail to reject the null hypothesis and conclude the variances are the same.

Answer:B) Fail to reject the null hypothesis and conclude no significant difference in the variances. Step-by-step explanation:Hello!You have two samples:Sample oneX₁: grade of an exam corrected by professor 1"n₁=10Sample mean X₁[bar]=179.3Standard deviation S₁=22.4Sample 2X₂: grade of an exam corrected by professor 2"n₂= 10Sample mean X₂[bar]= 282.1Standard deviation S₂= 12.0The objective of this test is to compare the population variances of the two populations. To test the variance ratio, you have to use an F-test.H₀: σ²₁/σ₂²=1H₁: σ²₁/σ₂²≠1α: 0.10F= (S₁²/S₂²)*(σ²₁/σ₂²) ~F[tex]_{(n1-1);(n2-1)}[/tex]The critical region is two-tailed, remember, the F distribution cumulates from 0 to + ∞ and presents asymmetry with positive asymptote. The upper critical value is rather simple to find, but for the low one you have to do a little convertion, since the F-tables give the cumulative frecuencies for 90% ≥ probabilities. (At least the ones I use)For the lower critical value the correction is the following:[tex]F_{(n1-1);(n2-1);\alpha/2} = \frac{1}{F_{(n2-1);(n1-1);1-\alpha/2 } }[/tex][tex]F_{(n1-1);(n2-1);(\alpha/2 )} = F_{9;9;0.05} = \frac{1}{F_{9;9;0.95} }[/tex] = [tex]\frac{1}{3.18}[/tex] = 0.31 Upper critical value:[tex]F_{(n1-1);(n2-1);1-\alpha/2} = F_{9;9;0.95} = 3.18[/tex]So, if [tex]F_{H0}[/tex] ≤ 0.31 or if [tex]F_{H0}[/tex] ≥ 3.18 the you reject the null hypothesis.If 0.31 < [tex]F_{H0}[/tex] < 3.18 the you don't reject the null hypothesis.[tex]F_{H0}= \frac{22.4}{12} * 1 = 1.867[/tex] With this value, the decision is to not reject the null hypothesis. This means that there is no difference between the population variances of the grading process of the two professors.I hope it helps!