Question

Using the six step process and Minitab program complete the test for independence:

Problem Definition: People of differing ages turn to a variety of media to get their news. A study indicated the various age groups receive their news from the media groups shown in the following contingency table.

Medium	Under 36 years of age	36-50 years of age	Above 50 years of age
Local TV	107	119	133
National TV	73	102	127
Radio	75	97	109
Local Newspaper	52	79	107
Internet	95	83	76

Set up your spreadsheet as shown below and do not enter row or column totals (do not include in report). Set alpha at .05

For full points the problem must be setup as follows:

1. Set up problem using six step process
2. Enter table data into Minitab using data shown in table above.
3. Run Chi square Stat>Tables>Cross tabulation and Chi square test two way table in worksheet.
4. Be sure not to enter row or column totals into your table as the outcome will include these values.
5. Test for independence between the media source and age.
6. State your conclusion using both the critical value/critical ratio technique and the pvalue
7. Examine the Minitab readout to make sure Cochran’s rule is observed.
8. If statistically significant at alpha .05 pick a medium and age and discuss whether the cell is over=represented or under-represented. Place this discussion in your interpretation section along with your interpretation of the overall outcome of the test.

Answer 1

The hypothesis being tested is:

H0: The media source and age are independent

Ha: The media source and age are dependent

The Minitab output is:

Medium		Under 36 years of age	36-50 years of age	Above 50 years of age	Total
Local TV	Observed	107	119	133	359
	Expected	100.64	120.17	138.19	359.00
	O - E	6.36	-1.17	-5.19	0.00
	(O - E)² / E	0.40	0.01	0.20	0.61
National TV	Observed	73	102	127	302
	Expected	84.66	101.09	116.25	302.00
	O - E	-11.66	0.91	10.75	0.00
	(O - E)² / E	1.61	0.01	0.99	2.61
Radio	Observed	75	97	109	281
	Expected	78.77	94.06	108.17	281.00
	O - E	-3.77	2.94	0.83	0.00
	(O - E)² / E	0.18	0.09	0.01	0.28
Local Newspaper	Observed	52	79	107	238
	Expected	66.72	79.67	91.62	238.00
	O - E	-14.72	-0.67	15.38	0.00
	(O - E)² / E	3.25	0.01	2.58	5.84
Internet	Observed	95	83	76	254
	Expected	71.21	85.02	97.77	254.00
	O - E	23.79	-2.02	-21.77	0.00
	(O - E)² / E	7.95	0.05	4.85	12.85
Total	Observed	402	480	552	1434
	Expected	402.00	480.00	552.00	1434.00
	O - E	0.00	0.00	0.00	0.00
	(O - E)² / E	13.39	0.17	8.63	22.18
		15.51	critical value
		22.18	chi-square
		8	df
		.0046	p-value

The p-value is 0.0046.

Since the p-value (0.0046) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that the media source and age are dependent.