Question

A study was conducted to find out the type of relationship between randomly selected15 Masters of Education Degree students scores in Research Methods and Statistics at Mount Kenya University during the 2012/2013 academic year. The results were as displayed in the table below.
Student    Research marks   Statistics marks
1   84   69
2   74   64
3   48   56
4   54   72
5   72   85
6   71   68
7   96   87
8   75   86
9   69   71
10   100   91
11   23   31
12   58   65
13   94   89
14   76   71
15   52   54

i)   State the null and alternative hypotheses.                    (2mks)
ii)   Calculate Pearson correlation coefficient between the two sets of data.         (4mks)
iii)   What is your overall conclusion regarding the nature of relationship between the two variables if r( df = 14) at ∝ =.01 significance level?           (2mks)  
iv)   Fit a line of best fit (linear regression) of Y on X                    (4mks)
v)   Estimate the Y score of a student who obtained 6 in X                 (1mk)  

Answer 1

(i) The hypothesis being tested is:

H₀: r = 0

H_a: r ≠ 0

(ii) r = 0.88

(iii) The critical value is 0.623.

Since 0.88 > 0.623, we can reject the null hypothesis.

Therefore, we can conclude that the relationship is significant.

(iv) y = 22.15 + 0.69*x

(v) y = 22.15 + 0.69*6 = 26.29

x	y	(x - xbar)	(y - ybar)	(x - xbar)*(y - ybar)	(x - xbar)²	(y - ybar)²
84	69	14.26667	-1.6	-22.8267	203.5378	2.56
74	64	4.266667	-6.6	-28.16	18.20444	43.56
48	56	-21.7333	-14.6	317.3067	472.3378	213.16
54	72	-15.7333	1.4	-22.0267	247.5378	1.96
72	85	2.266667	14.4	32.64	5.137778	207.36
71	68	1.266667	-2.6	-3.29333	1.604444	6.76
96	87	26.26667	16.4	430.7733	689.9378	268.96
75	86	5.266667	15.4	81.10667	27.73778	237.16
69	71	-0.73333	0.4	-0.29333	0.537778	0.16
100	91	30.26667	20.4	617.44	916.0711	416.16
23	31	-46.7333	-39.6	1850.64	2184.004	1568.16
58	65	-11.7333	-5.6	65.70667	137.6711	31.36
94	89	24.26667	18.4	446.5067	588.8711	338.56
76	71	6.266667	0.4	2.506667	39.27111	0.16
52	54	-17.7333	-16.6	294.3733	314.4711	275.56
xbar (Average)	ybar (Average)	Σ(x - xbar)	Σ(y - ybar)	Σ(x - xbar)*(y - ybar)	Σ(x - xbar)²	Σ(y - ybar)²
69.73333	70.6	0	8.53E-14	4062.4	5846.933	3611.6
r = Σ(x - xbar)*(y - ybar)/√(Σ(x - xbar)²*Σ(y - ybar)²)	0.88
b1 = Σ(x - xbar)*(y - ybar)/Σ(x - xbar)²	0.69
bo = ybar - b1*xbar	22.15

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