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 15% 21% 25% 17% 22% 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 280 401 504 354 461

(a) Show that it is appropriate to carry out a chi-square test using these data. Each expected value is ≥

(b) Test to determine whether the viewing shares in the current rating period differ from those in the last rating period at the .10 level of significance. (Round your answer to 3 decimal places.) χ2 χ 2 H0. Conclude viewing shares of the current rating period from those of the last.

Answer 1

(a)

Observed Frequencies:

WDUX (News)	280
WWTV (News)	401
WACO (Cheers Reruns)	504
WTJW (News)	354
Others	461

Expected Frequencies:

WDUX (News)	2000X15/100=300
WWTV (News)	2000X21/100=420
WACO (Cheers Reruns)	2000X25/100=500
WTJW (News)	2000X17/100=340
Others	2000X22/100=440

It is appropriate to carry out a chi -square test using these data, because expected value is 5.

(b)

Observed Frequencies:

Channel	Observed (O)	Expected (O)	(O - E)2/E
WDUX (News)	280	300	1.3333
WWTV (News)	401	420	0.8595
WACO (Cheers Reruns)	504	500	0.0320
WTJW (News)	354	340	0.5765
Others	461	440	1.0023
Total = =			3.8036

So,

Test statistic is:

= 3.804

= 0.10

ndf = 5- 1 = 4

From Table, critical value of = 7.779

Since the calculated value of = 3.804 is less than critical value of = 7.779, the difference is not significant. Fail to reject null hypothesis.

Conclusion:

The data do not support the claim that the viewing shares in the currect rating period differ from those in the last rating period.