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 wondered if a person’s age had arelationshipwith their desire to know or not know their true medical condition. Because of ethical constraints, the therapist asked visitors whether they would wish to be told if they had a terminal illness. What statistical test would you use and nine other pieces of information. The results are shown in the following table:

		Want to be informed	Do not want to be informed	Not sure
AGE
	<21	90	15	18
	22-35	56	24	19
	36-55	50	40	30
	>55	47	60	50

Pick 10

Level of data --

Parametric or nonparametric –

What statistical test would you perform?

Null hypothesis –

Research hypothesis – non directional –

Research hypothesis -- Directional–

What are the expected frequencies?

What alpha did you use and why?

How many degrees of freedom? Why? –

What is the calculated chi-square? (obtained) Did you use Yates? Why or why not? –

What is the critical value of chi-square? (Table F in Appendix C page 282)

Do you reject or accept the null?

What type of error would you have committed if you wrongly rejected or accepted?

Answer 1

Solution:

Chi-square test of independence

Level of data = nominal

This is a non-parametric test.

Here, we have to use chi square test for independence of two categorical variables.

Null hypothesis: H₀: A person’s age had no any relationship with their desire to know or not know their true medical condition.

Alternative hypothesis: H_a: A person’s age had a relationship with their desire to know or not know their true medical condition.

We assume level of significance = α = 0.05

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

E = row total * column total / Grand total

We are given

Number of rows = r = 4

Number of columns = c = 3

Degrees of freedom = df = (r – 1)*(c – 1) = 3*2 = 6

α = 0.05

Critical value = 12.59159

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies
	Column variable
Row variable	C1	C2	C3	Total
R1	90	15	18	123
R2	56	24	19	99
R3	50	40	30	120
R4	47	60	50	157
Total	243	139	117	499

Expected Frequencies
	Column variable
Row variable	C1	C2	C3	Total
R1	59.8978	34.26253	28.83968	123
R2	48.21042	27.57715	23.21242	99
R3	58.43687	33.42685	28.13627	120
R4	76.45491	43.73347	36.81162	157
Total	243	139	117	499

Calculations
(O - E)
30.1022	-19.2625	-10.8397
7.789579	-3.57715	-4.21242
-8.43687	6.573146	1.863727
-29.4549	16.26653	13.18838

(O - E)^2/E
15.12815	10.82947	4.074201
1.258598	0.464008	0.764441
1.218081	1.292561	0.123452
11.34776	6.050289	4.724955

(All calculations done by using excel)

Chi square = ∑[(O – E)^2/E] = 57.27596

P-value = 0.0000

(By using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that A person’s age had a relationship with their desire to know or not know their true medical condition.

Type I error we would have committed if we wrongly reject the null hypothesis.