Question

A statistics teacher predicts that the distribution of grades on the final exam will be 25% A's, 30% B's, 35% C's, and 10% D's. The actual distribution of grades on the final exam for 20 students is as follows. Using a 0.05 level of significance, is the statistics teacher's prediction correct? Number Grade of Students 16 P B un IC 16 D 3 Question 38 (2.63 points) Which test are you running? ANOVA One mean zetest c Chesquare Goodness of Fit One mean t-test Question 39 1263 points) BE han

Answer 1

A statistics teacher predicts that distribution of grades are 25% A's, 30% B's, 35% C's, 10% D's. So, we have to test that teacher's grade distribution fits the stated distribution.

Q. 38) -

We have to use chi-square test for goodness of fit, because this test is used to test that the given data is fit for the distribution.

While, ANOVA test used to test the significance of several means, one mean t-test & Z-test used to test the significance of single sample mean. Hence, we can't use these tests.

Option C is correct.

Q. 39) -

Null hypothesis - H₀ : teacher's grade distribution fits the stated distribution. 25% A's, 30% B's, 35% C's, 10% D's.

Option A is correct.

Q. 40) -

Alternative hypothesis - H₀ : teacher's grade distribution doea not fits the stated distribution. 25% A's, 30% B's, 35% C's, 10% D's.

Option B is correct.

Q. 41) -

Expected number of B's = N$\times$(probability of B) = 20$\times$0.3 = 6

Option A is correct.

Q.42) -

Test statistic -

$\chi ^{2} = \sum \frac{(O_{i}-E_{i})^2}{E_{i}}$

Observation table -

Grade	Number of students(oi)	Pi	Ei	(Oi-Ei)^2/Ei
A	6	0.25	5	0.2
B	5	0.3	6	0.1667
C	6	0.35	7	0.1429
D	3	0.1	2	0.5
Total	20	1	20	1.0096

Calculations -

$\chi ^{2} = \sum \frac{(O_{i}-E_{i})^2}{E_{i}} = 1.0096 \approx 1.01$

Q. 43) -

P-value = P($\chi ^{2}$ > $\chi ^{2} _{calc}$ ) = P($\chi ^{2}$ > 1.01) = 0.908

Q. 44) -

P-value > 0.05

So, we fail to reject the null hypothesis.

Option B is correct.

Q. 45) -

Conclusion -

At 0.05 level of significance, teacher's grade distribution fits the stated distribution. 25% A's, 30% B's, 35% C's, 10% D's.