COMM 214 FINAL MOCK EXAM COMPLETE QUESTIONS
AND 100% CORRECT ANSWERS
1. A hypothesis test is conducted at α = 0.05. If the p-value is 0.03,
what is the correct decision?
A. Reject H₀ ✓
B. Fail to reject H₀
C. Accept H₀
D. Conduct another test
Rationale: Since p-value (0.03) < α (0.05), we reject the null hypothesis.
The evidence is statistically significant.
2. What is the appropriate null hypothesis for testing if a car averages
more than 15 km/L?
A. H₀: μ < 15
B. H₀: μ = 15 ✓
C. H₀: μ > 15
D. H₀: μ ≠ 15
Rationale: The null hypothesis always contains the equality claim. The
alternative is Hₐ: μ > 15 (the manufacturer's claim).
3. In the car mileage problem, the test statistic is closest to:
A. 1.25
,B. 2.50 ✓
C. 3.75
D. 5.00
Rationale: z = (15.15 - 15)/(1.20/√400) = 0.15/0.06 = 2.50
4. For the car mileage test at α = 0.05, the critical value is:
A. 1.645 ✓
B. 1.960
C. 2.326
D. 2.576
Rationale: This is a right-tailed test (claim is "more than"), so the critical
value is z₀.₀₅ = 1.645.
5. The p-value for the car mileage test is approximately:
A. 0.0062 ✓
B. 0.0124
C. 0.4938
D. 0.9938
Rationale: z = 2.50, P(Z > 2.50) = 1 - 0.9938 = 0.0062.
6. A Type I error in the car mileage context would be:
A. Concluding the car averages more than 15 km/L when it does not ✓
B. Concluding the car does not average more than 15 km/L when it does
C. Concluding the car averages exactly 15 km/L
D. Failing to test enough cars
,Rationale: Type I error is rejecting a true null hypothesis. Here, rejecting
H₀ (μ = 15) when it's true means falsely claiming the car exceeds 15
km/L.
7. A Type II error in the car mileage context would be:
A. Concluding the car averages more than 15 km/L when it does not
B. Concluding the car does not average more than 15 km/L when it does
✓
C. Rejecting a false null hypothesis
D. Accepting a true null hypothesis
Rationale: Type II error is failing to reject a false null hypothesis. Here,
failing to conclude the car exceeds 15 km/L when it actually does.
8. If the sample size in the car test were increased from 400 to 1600,
the standard error would:
A. Increase by a factor of 2
B. Decrease by a factor of 2 ✓
C. Remain unchanged
D. Decrease by a factor of 4
Rationale: Standard error = σ/√n. If n quadruples, √n doubles, so
standard error is halved.
9. For the age-promotion study, what is the appropriate test?
A. Chi-square test of independence ✓
B. Z-test for proportions
, C. T-test for means
D. F-test for variances
Rationale: The data is categorical (age categories and promotion
status), and we're testing relationship between two categorical
variables.
10. In the age-promotion study, the degrees of freedom for the chi-
square test are:
A. 2
B. 3 ✓
C. 4
D. 8
Rationale: df = (r-1)(c-1) = (4-1)(2-1) = 3 × 1 = 3
11. The expected count for "Under 30 and Promoted" is closest to:
A. 7.5
B. 8.0 ✓
C. 9.0
D. 10.0
Rationale: Row total for Promoted = 9+29+32+10 = 80. Column total for
Under 30 = 9+41 = 50. Expected = (80×50)/250 = 16. Wait, recalc: Total
= 250, Row total = 80, Column total = 50, Expected = (80×50)/250 =
16.0. The provided options don't include 16. Actually checking the
table: Under 30 column total = 9+41 = 50, Promoted row total =
9+29+32+10 = 80, Expected = 80×50/250 = 16. None match. Let me
AND 100% CORRECT ANSWERS
1. A hypothesis test is conducted at α = 0.05. If the p-value is 0.03,
what is the correct decision?
A. Reject H₀ ✓
B. Fail to reject H₀
C. Accept H₀
D. Conduct another test
Rationale: Since p-value (0.03) < α (0.05), we reject the null hypothesis.
The evidence is statistically significant.
2. What is the appropriate null hypothesis for testing if a car averages
more than 15 km/L?
A. H₀: μ < 15
B. H₀: μ = 15 ✓
C. H₀: μ > 15
D. H₀: μ ≠ 15
Rationale: The null hypothesis always contains the equality claim. The
alternative is Hₐ: μ > 15 (the manufacturer's claim).
3. In the car mileage problem, the test statistic is closest to:
A. 1.25
,B. 2.50 ✓
C. 3.75
D. 5.00
Rationale: z = (15.15 - 15)/(1.20/√400) = 0.15/0.06 = 2.50
4. For the car mileage test at α = 0.05, the critical value is:
A. 1.645 ✓
B. 1.960
C. 2.326
D. 2.576
Rationale: This is a right-tailed test (claim is "more than"), so the critical
value is z₀.₀₅ = 1.645.
5. The p-value for the car mileage test is approximately:
A. 0.0062 ✓
B. 0.0124
C. 0.4938
D. 0.9938
Rationale: z = 2.50, P(Z > 2.50) = 1 - 0.9938 = 0.0062.
6. A Type I error in the car mileage context would be:
A. Concluding the car averages more than 15 km/L when it does not ✓
B. Concluding the car does not average more than 15 km/L when it does
C. Concluding the car averages exactly 15 km/L
D. Failing to test enough cars
,Rationale: Type I error is rejecting a true null hypothesis. Here, rejecting
H₀ (μ = 15) when it's true means falsely claiming the car exceeds 15
km/L.
7. A Type II error in the car mileage context would be:
A. Concluding the car averages more than 15 km/L when it does not
B. Concluding the car does not average more than 15 km/L when it does
✓
C. Rejecting a false null hypothesis
D. Accepting a true null hypothesis
Rationale: Type II error is failing to reject a false null hypothesis. Here,
failing to conclude the car exceeds 15 km/L when it actually does.
8. If the sample size in the car test were increased from 400 to 1600,
the standard error would:
A. Increase by a factor of 2
B. Decrease by a factor of 2 ✓
C. Remain unchanged
D. Decrease by a factor of 4
Rationale: Standard error = σ/√n. If n quadruples, √n doubles, so
standard error is halved.
9. For the age-promotion study, what is the appropriate test?
A. Chi-square test of independence ✓
B. Z-test for proportions
, C. T-test for means
D. F-test for variances
Rationale: The data is categorical (age categories and promotion
status), and we're testing relationship between two categorical
variables.
10. In the age-promotion study, the degrees of freedom for the chi-
square test are:
A. 2
B. 3 ✓
C. 4
D. 8
Rationale: df = (r-1)(c-1) = (4-1)(2-1) = 3 × 1 = 3
11. The expected count for "Under 30 and Promoted" is closest to:
A. 7.5
B. 8.0 ✓
C. 9.0
D. 10.0
Rationale: Row total for Promoted = 9+29+32+10 = 80. Column total for
Under 30 = 9+41 = 50. Expected = (80×50)/250 = 16. Wait, recalc: Total
= 250, Row total = 80, Column total = 50, Expected = (80×50)/250 =
16.0. The provided options don't include 16. Actually checking the
table: Under 30 column total = 9+41 = 50, Promoted row total =
9+29+32+10 = 80, Expected = 80×50/250 = 16. None match. Let me
