STAT 3013: Statistical Analysis of Exam Scores - Fall 2018

This assignment analyzes two sets of exam scores using various statistical methods. The analysis begins with the construction and comparison of boxplots for each exam to identify differences in score distributions and the presence of outliers. Histograms are then created to assess whether the scores follow a normal distribution. An F-test is performed to determine if the variances between the two exams are significantly different. A 95% confidence interval is constructed for the difference between the mean scores to compare the means of the two exams. Finally, a hypothesis test is conducted to determine if the mean score of one exam is significantly higher than the other. The results of each statistical test are presented with interpretations based on the data provided and the specified significance levels.

STAT 3013
Institution Name
Student Name
Date of Submission

1. Construction of the boxplots
When the two boxplots are compared its evident that the distributions of the scores do
differ. The median, minimum as well as the maximum point in each of the data set is
difference from the other.
The two data sets have no outliers as there is no indication from the boxplot
presented.
2. Histogram
Exam
1
bin
Frequenc
y
40 1
50 2
60 3
70 12
80 3
90 11
More 8

40 50 60 70 80 90 More
0
2
4
6
8
10
12
14
Histogram
Frequency
bin
Frequency
Exam
2
bin
Frequenc
y
60 1
70 5
80 11
90 11
More 10
60 70 80 90 More
0
2
4
6
8
10
12
Histogram
Frequency
bin
Frequency
From the histograms developed for the scores in exam 1 and 2 the two scores do not
appear to follow a normal distribution. This is because if we plot a line curve through
the centre of the pillars it does not produce a bell-shaped graph.
3. F-test

From the given data
Exam 1 Exam 2
32 55
45 62
50 63
56 66
58 67
60 67
61 71
61 73
63 74
64 74
64 74
65 75
66 76
67 77
67 77
68 78
69 80
69 82
72 83
76 84
78 85
81 86
83 86
83 87
85 87
86 88
87 89
87 90
88 91
89 91
90 92
90 93
91 93
92 94
92 94
93 95
93 97
94 99
96
98
The F-test was used to test the hypothesis

H0 :Var 1=Var 2
Vs
H1 :Var 1 ≠Var 2
From the excel output below the P value is 0.0135 which is less than 0.05. This falls
within the rejection region hence we reject the null hypothesis. We thereby conclude
that the variances of the two examinations are significantly difference.
F-Test Two-Sample for Variances
Exam 1 Exam 2
Mean 75.2250 81.4474
Variance 250.6404 120.3620
Observations 40 38
df 39 37
F 2.0824
P(F<=f) one-tail 0.0135
F Critical one-tail 1.7208
4. Confidence Interval

Exam 1 Exam 2
Mean 75.225 Mean 81.44736842
Standard Error 2.503199875 Standard Error 1.779725141
Median 77 Median 83.5
Mode 61 Mode 74
Standard Deviation 15.83162609 Standard Deviation 10.97096258
Sample Variance 250.6403846 Sample Variance 120.3620199
Kurtosis -0.196045775 Kurtosis -0.545489604
Skewness -0.576070143 Skewness -0.47377318
Range 66 Range 44
Minimum 32 Minimum 55
Maximum 98 Maximum 99
Sum 3009 Sum 3095
Count 40 Count 38
Confidence Level(95.0%) 5.063199659 Confidence Level(95.0%) 3.606065666
Mean variation
Exam 1 Exam 2
70.16180034 80.28819966 77.84130275 85.05343409
From the constructed confidence interval, it can be noticed that the value of the mean
does overlap for the two examinations. It can therefore be deduced that the mean of
the two exam scores are equal and neither exam have a higher mean than the other
5. Hypothesis test for the mean
Using the data

Exam 1 Exam 2
32 55
45 62
50 63
56 66
58 67
60 67
61 71
61 73
63 74
64 74
64 74
65 75
66 76
67 77
67 77
68 78
69 80
69 82
72 83
76 84
78 85
81 86
83 86
83 87
85 87
86 88
87 89
87 90
88 91
89 91
90 92
90 93
91 93
92 94
92 94
93 95
93 97
94 99
96
98
The test conducted is used to test the hypothesis
H0 : Mean1=Mean2
Vs

H1 : mean1<mean 2
t-Test: Two-Sample Assuming Unequal Variances
Exam 1 Exam 2
Mean 75.225 81.44736842
Variance 250.6403846 120.3620199
Observations 40 38
Hypothesized Mean Difference 0
df 70
t Stat -2.025913408
P(T<=t) one-tail 0.023293331
t Critical one-tail 2.380807482
P(T<=t) two-tail 0.046586662
t Critical two-tail 2.647904624
From the excel output we deduced the P value as 0.02329 which is greater than 0.01.
This does not fall in the rejection region. We thereby fail to reject the null hypothesis
and conclude that the mean of the two examinations have no significance difference.
We are 97.67% confidence that the means of the two exams show no statistical
variations.
References
Bowerman, B., O'Connell, R. & Murphree, E., 2017. Business Statistics in Practice: Using Data,
Modeling, and Analytics. 8th ed. s.l.:McGraw-Hill Education.