This question is a modified version In the study by Silver and Aiello, a seconda

Question

This question is a modified version

In the study by Silver and Aiello, a secondary objective was to determine if the frequency of falls was independent of wheelchair use. The following table gives the data for falls and wheelchair use among the subjects of the study.

           

Wheelchair Use

Yes

No

Fallers

62

121

Non-fallers

18

32

(1) Do these data provide sufficient evidence to warrant the conclusion that wheelchair use and falling are related? Let = .05. Use probability comparison test.

(2) Do these data provide sufficient evidence to warrant the conclusion that wheelchair use and falling are related? Let = .05. Use chi-square independence test.

(3) Are there any difference between two test results? If so, why do you think there it is? If not, which method do you think is better for this research? And, why?

Wheelchair Use

Yes

No

Fallers

62

121

Non-fallers

18

32

Explanation / Answer

(1) Here by probbility comparison test

pfallers (Wheel- chairs use)= 62/(62 + 18) = 0.775

pfallers (Now wheel chair use) = 121/ (121 + 32) = 0.791

Pooled estimate p = (62 + 121)/ (80 + 153) = 0.785

Here,

Test statistic

Z = (0.791 - 0.775)/ [0.785 * sqrt (1/n1 + 1/n2)] = 0.016 /[0.785 * 0.138] = 0.1477

So, Z < 1.96 (Z critical) so we can say that wheelchair use and falling are not correlated as proportion of fallers are same for both.

(2) CHi- square independence test.

Here the observed table

Expected table

Chi - square statistic

so

X2 = 0.783

here dF = 1 and alpha = 0.05

X20.05,1 = 3.8415

so here also we can say that both events are independent of each other.

(3 No, there is no difference between two test results. We thing the chi- square method is better as it entials and use all the cell of the contigency table.

Wheelchair Use Yes No Total Fallers 62 121 183 Non-fallers 18 32 50 Total 80 153 233

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