Question

An eigmeyear conort sludy of 3131 initially disease-free men and women (ages around 60 years or more) recorded the extent of physical activity the particpants engaged in and whether or not t were showing signs of "healthy aging" (British Journal of Sports Medicine, 2013) Here is the table of summarized counts at the conclusion of the study: Activity level Inactive Moderate Vigorous Total Yes 57 296 1194 203 556 2575 Healthy aging? 875 1078 No 506 Total 1490 3131 563

Answer 1

SOLUTION:

From given data,

O	Inactive	Moderate	Vigorous	Total
Yes	57	296	203	556
No	506	1194	875	2575
Total	563	1490	1078	3131

Calculate "Expected Value" for each entry:

Multiply each row total by each column total and divide by the overall total:

E = (Row total)(Column total) / Grand Total

E	Inactive	Moderate	Vigorous
Yes	563*556/3131 = 99.977004	1490*556/3131 =264.592782	1078*556/3131 =191.430214
No	563*2575/3131 =463.022996	1490*2575/3131 =1225.407218	1078*2575/3131 =886.569786

Subtract expected from actual, square it, then divide by expected:

$\chi ^{2}$ = $\Sigma (O-E)^{2} /E$

	Inactive	Moderate	Vigorous
Yes	(57-99.977004)2/99.977004 =18.474477	(296-264.592782)2/264.592782= 3.728043	(203-191.430214)2/191.430214=0.699262
No	(506-463.022996)2/463.022996=3.989052	(1194-1225.407218)2/1225.407218 = 0.804968	(875-886.569786)2/886.569786 = 0.150986

(i) the chi-square statistics

$\chi ^{2}$ = $\Sigma (O-E)^{2} /E$

= 18.474477 +3.728043+0.699262+3.989052+0.804968+0.150986

= 27.847

Degrees of Freedom

= (rows − 1) $\times$ (columns − 1)

DF = (2 − 1)(3 − 1) = 1×2 = 2

p- value at  $\alpha$ = 0.05 with   DF = 2

P-value = 0.0001

Critical value :

Critical value at  $\alpha$ = 0.05 with   DF = 2

$\chi _{0.05,2}^{2}$ = 5.991

Critical value = 5.991

Interpret the p-value:

Answer : first option is correct

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