Question

1. A radio executive considering a switch in his radio format collects the following data on the radio preferences of listeners in various age groups.

Radio preference	Age
	Young adult	Middle Age	Older adult
Music	14	10	3
News/talk	4	15	11
Sports	7	9	5

Test the null hypothesis that radio preference does not vary by age. Use a .05 confidence level for your test and report (a) the critical value of the test statistic, (b) the obtained value of the test statistic, (c) your decision about the null hypothesis, and (d) an interpretation of the meaning of your decision.

Answer 1

Observed Frequencies (E)	Young Adult	Middle Age	Older Adult	Total
Music	14	10	3	27
News/Talk	4	15	11	30
Sports	7	9	5	21
Total	25	34	19	78

The null and alternative hypotheses are    
Ho : Age and radio preferences are independent    
Ha : Age and radio preferences are not independent    
     
a) α = 0.05    
We use Chi Square goodness of fit test    
Degrees of freedom = df = (number of rows - 1) * (number of columns - 1)    
Degrees of freedom = df = 4    
     
We find Critical value of the test statistic using Excel Function CHISQ.INV.RT    
χ² critical = CHISQ.INV.RT(0.05, 4)    
χ² critical = 9.4877    
     
b) Grand Total of frequencies = 78    
To find Expected Frequencies    
Expected Frequencies = (Row Total * Column Total)/Grand Total    

Expected Frequencies (E)	Young Adult	Middle Age	Older Adult
Music	8.6538	11.7692	6.5769
News/Talk	9.6154	13.0769	7.3077
Sports	6.7308	9.1538	5.1154

Following table gives the value of (Observed - Expected)² / Expected

	Young Adult	Middle Age	Older Adult
Music	3.3027	0.266	1.9453
News/Talk	3.2794	0.2828	1.8656
Sports	0.0108	0.0026	0.0026

Chi Square Value = ∑[(Observed - Expected)² / Expected]    
Test Statistic χ² = 10.9578    
     
c) Decision :    
10.9578 > 9.4877    
Test Statistic χ² > χ² critical    
Hence we REJECT Ho    
     
     
Interpretation:    
There exists enough statistical evidence at α = 0.05 to show that     
age and radio preferences are not independent