A family therapist in a hospital wanted to know if patients with a terminal illn

Question

A family therapist in a hospital wanted to know if patients with a terminal illness wanted to be informed of their true medical condition. The therapist also wondered if a person’s age has an effect on their attitude. Because of ethical constraints the therapist asked a healthy sample of subjects who were visitors to the hospital whether they would wish to be told if they had a terminal illness. The age of the respondents was also recorded. The results are shown in the table below.

1) Give the appropriate hypotheses.

2) Give the values of fe for the three cells pertaining to people who are between 22 and 35 years old.

3) How many df are there? Either give 2-crit or the p-value of the test.

4) What do you conclude?

Attitude AGE Want to be informed Did not want to be informed Not sure < 21 90 16 18 124 22-35 56 24 19 99 36-55 50 40 30 120 > 55 47 60 50 157 243 140 117 500

Explanation / Answer

here we use 2 test of independent of attributes with

and  2=sum((O-E)2/E) with (r-1)(c-1) df

(1) null hypothesis H0: age has not an effect on their attitude

alternate hypothesis H1: age has an effect on their attitude

(2)

BOLD CELSL PERTAINING TO REQUIRED WHO ARE BETWEEN 22 AND 35 ARES

(3) df=(4-1)*(3-1)=6, calculated chi-square=55.92

and p-value is less than 0.0001 and critical chi-squrar(0.05,6)=12.59

alpha is not given so let here level of significance is alpha=0.05

(4) since p-value is less than alpha=0.05 and critial chi-square is less than calculated chi-square, so we fail to accept H0 and conclude that age has an effect on their attitude

observed(O) Expected(E) E observed(O) (O-E) (O-E)2/E E(90)= 243*124/500 60.26 90 29.74 14.6726 E(56)= 243*99/500 48.11 56 7.89 1.292534 E(50)= 243*120/500 58.32 50 -8.32 1.186941 E(47)= 243*157/500 76.30 47 -29.30 11.25275 E(16)= 140*124/500 34.72 16 -18.72 10.09327 E(24)= 140*99/500 27.72 24 -3.72 0.499221 E(40)= 140*120/500 33.60 40 6.40 1.219048 E(60)= 140*157/500 43.96 60 16.04 5.85263 E(18)= 117*124/500 29.02 18 -11.02 4.182253 E(19)= 117*99/500 23.17 19 -4.17 0.749182 E(30)= 117*120/500 28.08 30 1.92 0.131282 E(50)= 117*157/500 36.74 50 13.26 4.787431 sum= 500.00 500.00 0.00 55.92

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