DATA ANALYSIS
Problem 1
Answer the question: Does practice time make a difference? Support your answer with your findings.
The ANOVA single factor test was used to determine whether there was a difference in the swimming time by practice time. ANOVA was calculated using the following steps:
i) Present the data in three columns by practice time as shown:
less than 15 (1)
15-25 (2)
greater than 25(3)
58.7
64.4
68
55.3
55.8
65.9
61.8
58.7
54.7
49.5
54.7
53.6
64.5
52.7
58.7
61
67.8
58.7
65.7
61.6
65.7
51.4
58.7
66.5
53.6
54.6
56.7
59
51.5
55.4
54.7
51.5
61.4
54.8
56.9
57.2
ii) State the null and alternative hypotheses
Ho: all means are equal
HA: all means are not equal
iii) Run a single-factor ANOVA using the data analysis tool on Excel. The results are presented below:
Anova: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
Column 1
10
580.5
58.05
29.87833
Column 2
13
753.5
57.96154
22.68423
Column 3
13
767.4
59.03077
31.07897
ANOVA
Source of Variation
SS
df
MS
F
P-value
F crit
Between Groups
8.868760684
2
4.43438
0.160092
0.852723
3.284918
Within Groups
914.0634615
33
27.69889
Total
922.9322222
35
Implication: The F statistic (0.16) is less than the F critical (3.29). Further, the F-statistic is yields a p-value of 0.853, which is greater than p?0.05, implying that the differences in means is not significant at the .05 level of significance and hence, we accept the null hypothesis. This implies that there is no difference in swimming time by the number of practice hours. As such, practice time does not make a difference in swimming time.
Problem 2
1. Compute the correlation between motivation (x) and GPA (y).
Correlation indicates the strength and direction of the association between two variables. It is computed using the formula:
The steps in computing the correlation...
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