A researcher measures the relationship between Internet use (hours per week) and

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.

(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.

Internet Use Social Interaction X Y 6 5 9 6 6 6 6 7 12 4 3 7 3 3 6 4 1 10 10 4

Explanation / Answer

(a) Here to compute the pearson correlation coefficient , the formula is

r =  [nxy - xy] / sqrt[ (n(x2 - (x)2) (n(y)2 - y2)]

so, we have to calculate these values

r = [10 * 314 - 62 * 56]/ sqrt [ (10 * 488 - 622) * (10 * 352 - 562) ]

r = -332/ sqrt[ 1036 * 384]

r = -332/ 630.73 = -0.526

(b) coefficient of determination.  

R2 = (-0.526)2 = 0.277

(c) Two tailed test at 0.05 level of significance.

t = r sqrt (n-2)/ (1 -r2)]

t = -0.526 * sqrt [ (10-2)/ (1 - 0.277)]

t = -0.526 * sqrt [8/ 0.723]

t = - 1.75

so at significance level alpha = 0.05

tcr = 2.306

so we here l t l > critical value of t so we shalln't reject the null hypothesis and say that it is not significant.

Internet Use Social Interaction X Y X^2 Y^2 XY 6 5 36 25 30 9 6 81 36 54 6 6 36 36 36 6 7 36 49 42 12 4 144 16 48 3 7 9 49 21 3 3 9 9 9 6 4 36 16 24 1 10 1 100 10 10 4 100 16 40 Sum 62 56 488 352 314

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