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 2 4 7 5 7 6 12 4 4 6 4 3 5 5 2 10 3 (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 O Retain the null hypothesis. Reject the null hypothesis.

Answer 1

We need to find the sums ∑x , ∑y, ∑x*y, ∑x² and ∑y²

2 х у 4 x? 49 64 7 y? 16 49 xty 28 8 7 56 5 7 25 49 35 6 6 36 36 36 16 12 4 144 48 4 6 16 36 24 4 3 16 9 12 5 5 25 25 25 2 4 100 20 10 3 11 121 9 345 33 317 Total 64 55 500

∑x = 64 , ∑y = 55 , ∑x*y = 317 , ∑x² = 500 and ∑y² = 345

a ) Correlation coefficient r :

r = $\frac{n\sum x*y-\sum x*\sum y}{\sqrt{[n\sum x^2-(\sum x)^2 ]*[n\sum y^2-(\sum y)^2]}}$

r = $\frac{(10*317)-(64*55)}{\sqrt{[(10*500)-(64*64) ]*[(10*345)-(55*55)]}}$

r = -350 / 619.8387

r = -0.565

b) Coefficient of determination ( $r^2$ ) = 0.319

c) r = -0.565 , | r | = 0.565

We have n =10 , α = 0.05

n 8 9 H 10 2-tailed testing a = .1 a = '05 a = .01 0.621 0.707 0.834 0.582 0.666 0.798 0.549 0.632 0.765

Therefore critical value is 0.632

As | r | is less than 0.632 , We retain the null hypothesis.