Data Analysis using Correlation and ANOVA

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 coefficient are:

i) Obtain the mean (average) for both motivation (x) and GPA (y)

Motivation Average x? = 5.933

GPA Average y? = 2.833

ii) Calculate ? (x-x? ) (y- y?) = 16.77

iii) Calculate ? (x-x? ) 2 = 151.87

iv) Compute ? (y- y?) 2 = 9.83

v) Compute ?(? (x-x? ) 2 ? (y- y?) 2) = 151.87 (9.83) = ?1,492 = 38.63

vi) Compute r = 16.77/38.63 = 0.43

2. Test for the significance of the correlation coefficient at the .05 level using a two-tailed test.

The results of the two-tailed t-test are presented below:

t-Test: Two-Sample Assuming Equal Variances

 

Variable 1

Variable 2

Mean

5.933333333

2.833333

Variance

5.236781609

0.338851

Observations

30

30

Pooled Variance

2.787816092

Hypothesized Mean Difference

0

df

58

t Stat

7.190767768

P(T

7.00622E-10

t Critical one-tail

1.671552762

P(T

1.40124E-09

t Critical two-tail

2.001717484

 

The null hypothesis is that there is zero correlation between the two variables. The p-value for the two-tail is 1.4, which is less than the t-critical of 2.00, we accept the null hypothesis that the correlation coefficient is not significant at the .05 significance level.

3. True or false? The more highly you are motivated, the more you will study. Which did you select and why?