BSB123 Data Analysis

BSB123 Data Analysis
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Task 1
(a) A side-by-side box plot of result is in Figure 1. (Lem, Onghena, Verschaffel, & Van Dooren, 2013).
Figure 1: Distribution of results for males and females
The median of results for females is noted to be at 60, and the distribution of results is noted to be
almost normally distributed with a slight negative skewness implying presence of few girls with
higher results. Third quartile was around 70 signifying that score above which the top 25% females
scored. Hence the mid 50% females scored between 50 and 70.
Median of results for males is 55, and the distribution of results is noted to be almost normally
distributed with a slight negative skewness implying presence of some males with higher results.
Third quartile was around 65 signifying that score above which the top 25% males scored. Hence the
mid 50% males scored between 50 and 65. Therefore, on an average females are observed to fare
better compared to males (Devore, & Berk, 2012).
(b) Normality and independence was assumed. Homogeneity of the dependent variable was checked
using an F-test (Kim, 2015).
H 0: ( μM =μF ) : Average results for males and females are statistically same:
H 0: ( μF ≠μM ) : Average result for females is significantly different than that of the males:
The test is a two tailed test since the difference between the average results is checked.
Significance level: 5%
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Test statistic:
t = μF−μM
SE ( diff ) =3 . 32
and p < 0.05.
Decision: The null hypothesis is rejected on the basis of p-value.
Table 1: Excel output of the t-test for results between genders
(c) Normality and independence was assumed. Homogeneity of the dependent variable was checked
using an F-test.
H 0: ( μN =μY ) : Average results for both yes and no relationship statuses are statistically same:
H 0: ( μY < μN ) : Average result of students in a romantic relation is significantly lower than that
of the students with no relationship status:
This is the case of a left tailed test.
Significance level: 5%
Test statistic:
t = μN−μY
SE ( diff ) =2 .22
and p < 0.05.
Decision: The null hypothesis is rejected on the basis of p-value.
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Table 2: Excel output of the t-test for results between two relationship statuses
Task 2
(a) A correlation matrix of Result, Age, Lectures, and Tutorials has been constructed and presented in
Table 5. Very low and negative pairwise correlations between Result and other three variables can
be identified from Table 5. Hence, Age, Lectures, and Tutorials are predictors of Result.
Table 3: Correlation matrix of Result, Age, Lectures, and Tutorials
Correlation
Result
Age Lectures Tutorials
Result 1
Age -0.107 1
Lectures -0.092 0.152 1
Tutorials -0.084 0.147 0.946 1
(b)
(i)
Table 4: Simple regression for result on age
Regression equation is Results = 86.14 – 1.42 * Age
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