Question

A researcher measures the relationship between Internet use (hours per week) and social interaction (hours per week) in a sample of 10 students. The following table lists the hypothetical results of this study.

Internet Use	Social Interaction
X	Y
7	5
9	5
5	8
6	7
13	6
5	7
3	3
5	5
1	10
12	2

(a) Compute the Pearson correlation coefficient. (Round your answer to three decimal places.)


(b) Compute the coefficient of determination. (Round your answer to three decimal places.)


(c) Using a two-tailed test at a 0.05 level of significance, state the decision to retain or reject the null hypothesis.

Retain the null hypothesis. Reject the null hypothesis.    

Answer 1

Answer:

Given Data,

A researcher measures the relationship between Internet use (hours per week) and social interaction (hours per week) in a sample of 10 students.

The following table lists the hypothetical results of this study.

Internet Use	Social Interaction
X	Y
7	5
9	5
5	8
6	7
13	6
5	7
3	3
5	5
1	10
12	2

Let the table can be modified as

Internet Use	Social Interaction
X	Y	XY	X2	Y2
7	5	35	49	25
9	5	45	81	25
5	8	40	25	64
6	7	42	36	49
13	6	78	169	36
5	7	35	25	49
3	3	9	9	9
5	5	25	25	25
1	10	10	1	100
12	2	24	144	4
X=66	Y=58	XY=343	X2=564	Y2=386

(a). To find the Pearson correlation coefficient:

Pearson correlation coefficient

Here n=10

Therefore Pearson Correlation=-0.498

(b).The coefficient of determination:

The coefficient of dermination=

Therefore The Coefficient of dermination=0.248

(c).

Null Hypothesis H₀ : (Two-tailed test)

Alternative Hypothesis H₀ :

Therefore Test-Statistic(t)-1.873

P-value=0.216 (Since, Two-tailed test)

Given the significance level =0.05

P-value>significance level

i.e, 0.216>0.05

Fail to reject the null hypothesis(H₀).

Retain the null hypothesis.