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 29%, CBS 28%, NBC 25%, and independents 18%. A sample of 300 homes two weeks after a Saturday night schedule revision yielded the following viewing audience data: ABC 95 homes, CBS 70 homes, NBC 89 homes, and independents 46 homes.

Test with α = 0.05 to determine whether the viewing audience proportions changed.

State the null and alternative hypotheses.

H₀: p_ABC = 0.29, p_CBS = 0.28, p_NBC = 0.25, p_IND = 0.18
H_a: The proportions are not p_ABC = 0.29, p_CBS = 0.28, p_NBC = 0.25, p_IND = 0.18.H₀: p_ABC = 0.29, p_CBS = 0.28, p_NBC = 0.25, p_IND = 0.18
H_a: p_ABC ≠ 0.29, p_CBS ≠ 0.28, p_NBC ≠ 0.25, p_IND ≠ 0.18     H₀: p_ABC ≠ 0.29, p_CBS ≠ 0.28, p_NBC ≠ 0.25, p_IND ≠ 0.18
H_a: p_ABC = 0.29, p_CBS = 0.28, p_NBC = 0.25, p_IND = 0.18H₀: The proportions are not p_ABC = 0.29, p_CBS = 0.28, p_NBC = 0.25, p_IND = 0.18.
H_a: p_ABC = 0.29, p_CBS = 0.28, p_NBC = 0.25, p_IND = 0.18

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.)

p-value =

State your conclusion.

Reject H₀. There has been a significant change in the viewing audience proportions.

Do not reject H₀. There has been a significant change in the viewing audience proportions.     

Reject H₀. There has not been a significant change in the viewing audience proportions.

Do not reject H₀. There has not been a significant change in the viewing audience proportions.

Answer 1

H0: pABC = 0.29, pCBS = 0.28, pNBC = 0.25, pIND = 0.18
Ha: The proportions are not pABC = 0.29, pCBS = 0.28, pNBC = 0.25, pIND = 0.18.

observed frequencey, O	expected proportion	expected frequency,E	(O-E)		(O-E)²/E
95	0.290	87.00	8.00		0.736
70	0.280	84.00	-14.00		2.333
89	0.250	75.00	14.00		2.613
46	0.180	54.00	-8.00		1.185

chi square test statistic,X² = Σ(O-E)²/E =   6.867              
                  
level of significance, α=   0.05              
Degree of freedom=k-1=   4   -   1   =   3
                  
P value =   0.0762 [ excel function: =chisq.dist.rt(test-stat,df) ]          
Decision: P value >α , Do not reject Ho                  
Do not reject H0. There has not been a significant change in the viewing audience proportions.