Question

A radio executive considering a switch in his station’s format collects data on the radio preference of various age groups of listeners.  He wants to know if there is a relationshipbetween radio format preferences by age group.  

	Young Adult	Middle Age	Older Adult
Radio Format
Music	14	10	3
News-talk	4	15	11
Sports	7	9	5

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?

Do you reject or accept the null?

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

Answer 1

Level of data = nominal

This is a non-parametric test.

Chi-square test of independence

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

Null hypothesis: H₀: There is no relationship between radio format preferences by age group.

Alternative hypothesis: H_a: There is a relationship between radio format preferences by age group.

We assume the 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 = 3

Number of columns = c = 3

Degrees of freedom = df = (r – 1)*(c – 1) = 2*2 = 4

α = 0.05

Critical value = 9.487729

(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	14	10	3	27
R2	4	15	11	30
R3	7	9	5	21
Total	25	34	19	78

Expected Frequencies
	Column variable
Row variable	C1	C2	C3	Total
R1	8.653846	11.76923	6.576923	27
R2	9.615385	13.07692	7.307692	30
R3	6.730769	9.153846	5.115385	21
Total	25	34	19	78

Calculations
(O - E)
5.346154	-1.76923	-3.57692
-5.61538	1.923077	3.692308
0.269231	-0.15385	-0.11538

(O - E)^2/E
3.302735	0.265963	1.945344
3.279385	0.282805	1.865587
0.010769	0.002586	0.002603

Chi square = ∑[(O – E)^2/E] = 10.95778

P-value = 0.027043

(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 there is a relationship between radio format preferences by age group.

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