Question

During the first 13 weeks of the television season, the Saturday evening 8 p.m. to 9 p.m. audience proportions were recorded as ABC 31%, CBS 26%, NBC 27%, and independents 16%. A sample of 300 homes two weeks after a Saturday night schedule revision yielded the following viewing audience data: ABC 93 homes, CBS 72 homes, NBC 87 homes, and independents 48 homes. Test with α = 0.05 to determine whether the viewing audience proportions changed.

State the null and alternative hypotheses

Find the value of the test statistic. (Round your answer to three decimal places.)

Find the p-value. (Round your answer to four decimal places.)

State your conclusion.

Answer 1

Category	Observed Frequency (O)	Proportion, p	Expected Frequency (E)	(O-E)²/E
ABC	93	0.31	300 * 0.31 = 93	(93 - 93)²/93 = 0
CBS	72	0.26	300 * 0.26 = 78	(72 - 78)²/78 = 0.4615
NBC	87	0.27	300 * 0.27 = 81	(87 - 81)²/81 = 0.4444
Independent	48	0.16	300 * 0.16 = 48	(48 - 48)²/48 = 0
Total	300	1.00	300	0.9060

Null and Alternative hypothesis:

Ho: Proportions are same.

H1: Proportions are different.

Test statistic:

χ² = ∑ ((O-E)²/E) = 0.9060

df = n-1 = 3

p-value = CHISQ.DIST.RT(0.906, 3) = 0.824

Decision:

p-value > α, Do not reject the null hypothesis

There is not enough evidence to conclude that the viewing audience proportions changed at 0.05 significance level.