Data Analysis: Hypothesis Testing, ANOVA, Correlation, Regression

Data Analysis
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Question 7
a) Dependent variable: TMA scores
Independent variables: Three courses at SUSS
Null hypothesis:
H 0 : ( μ1=μ2=μ3 )
: Average TMA scores are same for three courses at SUSS.
Alternate hypothesis: All three average TMA scores are not same for the courses at SUSS.
b) The missing values are calculated as below,
i) SST = SSB + SSW = 3568.85 + 18467.34 = 22036.19
ii) Here, N =30 (total number of observations), k = number of groups
df W
=degrees of freedom within the group = N – k = 30 – 3 = 27
iii) Total degrees of freedom =
df T =df W + df B
= 27 + 2 = 29
iv) Mean Square within = MSW = SSW / dfW = 18467.34 / 27 = 683.97
v) F = MSB / MSW = 1784.42 / 683.97 = 2.61
c) Conclusion: Critical F-value at 5% level of significance for
df B=2
and
df W =27
is evaluated
as,
F ( 2 , 27 , 0 .05 ) =3 .35
. Calculated F-statistics = 2.61 < F-critical = 3.35.
Hence, the null hypothesis is failed to get rejected. Therefore, it can be concluded that there is no
significant difference in average TMA scores between the students from three courses at SUSS.
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d) ANOVA merely indicates existence of difference between the groups, but, does not indicate
the measure of difference between the groups. Effect size in one-way ANOVA indicates the
measure of difference between the groups, provided there is any difference.
Effect size is calculated as
η2
=SSB / SST = 3568.85 / 22036.19 = 0.162
The value of eta-square implied that only 16.2% variances in TMA sores can be explained by
three courses in SUSS. Hence, there is very less (insignificant) practical significance of the
differences in average TMA scores between the three courses.
Question 8
a) i) As the swimming coach wants to see improvement in average swimming timings, a right
tailed (one-tailed) test will be appropriate.
A two-tailed test will yield either significant difference in average swimming timings or the
difference will be statistically insignificant. In that case the result might be significant for
improved or worsened average swimming timing. Therefore, one-tailed test will provide
appropriate answer, whether the new timings have significantly improved.
ii) Null hypothesis:
H 0 : ( μ=μ0 )
=New average swimming time under new method is equal to
the old average swimming time.
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Alternate hypothesis:
HA : ( μ> μ0 )
=New average swimming time under new method is
significantly greater than old average swimming time.
iii) Mean =
X
¿
=∑ X
n
=69 . 49+70 . 34+ 63. 51+71. 30+70 . 19+67 .18 +72. 27+62 . 13+66 . 03+68 .23+71 . 33+69 . 45
12
=68 . 45
Sample standard deviation =
s= √ ( 69 . 49−68 . 45 ) 2+ ( 70 .34−68 . 45 ) 2+.. .+ ( 71 .33−68 . 45 ) 2+ ( 69. 45−68 . 45 ) 2
11 =3. 18
Population standard deviation =
σ =3 .2
Standard error =
σ X
¿ = σ
√ n = 3. 2
√ 12 =0 . 92
The test statistics is evaluated as
Zcal= X
¿
−μ
σ x
¿
=68 . 45−69 . 8
0 . 92 =−1. 47
The p-value was evaluated as
p=P ( Z>−1. 47 ) =P ( Z <1 . 47 )
= 0.929
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