Question

Last rating period, the percentages of viewers watching several channels between 11 p.m. and 11:30 p.m. in a major TV market were as follows:

WDUX (News)	WWTY (News)	WACO (Cheers Reruns)	WTJW (News)	Others
13%	24%	22%	16%	25%

Suppose that in the current rating period, a survey of 2,000 viewers gives the following frequencies:

WDUX (News)	WWTY (News)	WACO (Cheers Reruns)	WTJW (News)	Others
326	500	551	323	300

(a) Show that it is appropriate to carry out a chi-square test using these data and calculate the value of the test statistic.

Each expected value is ≥        

χ2χ2        

Answer 1

Null hypothesis : H_o : Current rating period viewer watching frequency  matches with the last rating period percentages

Alternate hypothesis: H_a : Current rating period viewer watching frequency does not match  with the last rating period percentages

Chi-square goodness of fit is appropriate for testing this hypothesis

Given : Observed Frequency :O

	O: Observed Frequency
WDUX(News)	326
WWTY(News)	500
WACO(Cheers Rerun	551
WTJW(News)	323
Others	300
Total	2000

Expected frequency for a given channel = Channel's last period percentage of 2000

i.e

	Last rating period , Percentage	E:Expected Frequency	E
WDUX(News)	13%	13% of 2000	260
WWTY(News)	24%	24% of 2000	480
WACO(Cheers Rerun	22%	22% of 2000	440
WTJW(News)	16%	16% of 2000	320
Others	25%	25% of 2000	500

	O	E	O-E	(O-E)2	(O-E)2/E
WDUX(News)	326	260	66	4356	16.75385
WWTY(News)	500	480	20	400	0.833333
WACO(Cheers Rerun	551	440	111	12321	28.00227
WTJW(News)	323	320	3	9	0.028125
Others	300	500	-200	40000	80
				Total	125.6176

value of the test statistic:

Value of the test statistic = 125.6176

------------------------------------------------

Degrees of freedom = 5-1 =4

For 4 degrees of freedom =

As p-value is less than the level of significance reject the null hypothesis.

There is sufficient evidence to conclude that Current rating period viewer watching frequency does not match  with the last rating period percentages