Question 1194707: A lecturer in a university wants to investigate the differences in students’ marks with the
different majors from School A, School B, School C and School D for MSG 162 subject.
It was thought that students with four different majors might result in different grades for
MSG 162. The lecturer examined exam scores for a sample of four students each from the
school that participated in the MSG 162 class
The following are the MSG 162 exam results for the students:
A B C D
85 67 88 77
79 75 79 56
76 72 85 60
83 80 82 70
Assume that all assumptions to perform an ANOVA test are satisfied.
a) Test the hypothesis that there is no difference between exam scores of students from
four schools at the significance level, α = 0.05. Interpret the results of the hypothesis
testing.
b) Determine the level of significance (p-value) of the test results.
c) Determine a 95% interval estimate for the students marks from school B.
Answer by proyaop(69)   (Show Source):
You can put this solution on YOUR website! **a) Hypothesis Testing**
* **Null Hypothesis (H0):** There is no significant difference in the mean exam scores of students from the four schools (μA = μB = μC = μD).
* **Alternative Hypothesis (H1):** At least one school's mean exam score is significantly different from the others.
**1. Calculate the necessary statistics:**
* **Calculate the mean and variance for each group:**
* School A: Mean = 80.75, Variance = 8.6875
* School B: Mean = 73.50, Variance = 10.1250
* School C: Mean = 83.50, Variance = 4.6875
* School D: Mean = 65.75, Variance = 66.1875
* **Calculate the overall mean:**
* Overall Mean = (Sum of all scores) / Total number of scores = 75.875
* **Calculate the between-groups sum of squares (SSB):**
* SSB = Σ(n_i * (mean_i - overall_mean)^2)
* SSB = 4 * ((80.75 - 75.875)^2 + (73.50 - 75.875)^2 + (83.50 - 75.875)^2 + (65.75 - 75.875)^2)
* SSB = 4 * (23.890625 + 5.421875 + 58.007813 + 100.15625)
* SSB = 752.5
* **Calculate the within-groups sum of squares (SSW):**
* SSW = Σ(n_i - 1) * variance_i
* SSW = (4 - 1) * (8.6875 + 10.1250 + 4.6875 + 66.1875)
* SSW = 246
* **Calculate the degrees of freedom:**
* Between-groups degrees of freedom (dfB) = k - 1 = 4 - 1 = 3
* Within-groups degrees of freedom (dfW) = N - k = 16 - 4 = 12
* **Calculate the mean squares:**
* Mean Square Between (MSB) = SSB / dfB = 752.5 / 3 = 250.83
* Mean Square Within (MSW) = SSW / dfW = 246 / 12 = 20.5
* **Calculate the F-statistic:**
* F = MSB / MSW = 250.83 / 20.5 = 12.23
**2. Determine the critical value:**
* Using an F-distribution table with dfB = 3 and dfW = 12, and α = 0.05, the critical value is approximately 3.49.
**3. Make a decision:**
* Since the calculated F-statistic (12.23) is greater than the critical value (3.49), we reject the null hypothesis.
**Interpretation:**
* There is sufficient evidence at the 0.05 significance level to conclude that there is a significant difference in the mean exam scores of students from the four schools.
**b) Determine the p-value:**
* Using statistical software (like R or Python), we can find the exact p-value associated with the calculated F-statistic (12.23) and the degrees of freedom (dfB = 3, dfW = 12).
* The p-value will be very small (likely less than 0.001).
**c) 95% Confidence Interval for the Mean of School B:**
* **Calculate the standard error of the mean for School B:**
* Standard Error (SE) = √(variance_B / n_B) = √(10.1250 / 4) = 1.586
* **Find the critical t-value:**
* For a 95% confidence interval and df = n_B - 1 = 3, the critical t-value (from a t-distribution table) is approximately 3.182.
* **Calculate the margin of error:**
* Margin of Error = t_critical * SE = 3.182 * 1.586 = 5.05
* **Calculate the confidence interval:**
* Lower limit = Mean_B - Margin of Error = 73.50 - 5.05 = 68.45
* Upper limit = Mean_B + Margin of Error = 73.50 + 5.05 = 78.55
* **95% Confidence Interval for School B:** (68.45, 78.55)
**Interpretation:**
We are 95% confident that the true mean exam score for students from School B lies between 68.45 and 78.55.
**Note:** This analysis assumes that the assumptions for ANOVA are met, including normality of the data within each group and homogeneity of variances.
I hope this comprehensive analysis is helpful!
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