During the first 13 weeks of the television season, the Saturday evening 8:00 p.m. to 9:00 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 97 homes, CBS 68 homes, NBC 91 homes, and independents 44 homes. Test with = .05 to determine whether the viewing audience proportions changed.

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batolisis batolisis · Answer 1 · 2021-05-24T16:04:21+02:00

Answer:

There is no change in the proportion of the viewing audience

Step-by-step explanation:

Aim : Determine if the viewing audience proportions changed

Recorded audience proportions : ABC 31%, CBS 26%, NBC 27%, and independents 16%

sample size : 300

significance level ( ∝ ) = 0.05

Results yielding from sample size : ABC 97 homes, CBS 68 homes, NBC 91 homes, and independents 44 homes

State the hypothesis of this sampling

H0 : p1 = 0.31 or p2 = 0.26 or p3 = 0.27 or p4 = 0.16

Ha : at least pi ≠ 0.31 , 0.26 , 0.27 , 0.16

next determine the test statistic

X^2 = ∑ ( Oi - Ei)^2 / Ei ---------- ( 1 )

For ABC

Oi = 97 , Ei = 300* 0.31 = 93 , ( Oi - Ei)^2 = 16 , ( Oi - Ei)^2 / Ei = 16 / 93 = 0.17

For CBS

Oi = 68 , Ei = 300 * 0.26 = 78, ( Oi - Ei)^2 = 100, ( Oi - Ei)^2 / Ei = 100 / 78 = 1.28

For NBC

Oi = 91 , Ei = 300 * 0.27 = 81 , ( Oi - Ei)^2 = 100, ( Oi - Ei)^2 / Ei = 100/81 = 1.23

Independent

Oi = 44 , Ei = 300*0.16 = 48, ( Oi - Ei )^2 = 16 , ( Oi - Ei)^2 / Ei = 16 /48 = 0.33

back to equation 1

X^2 = ( 0.17 + 1.28 + 1.23 + 0.33 ) = 3.01

hence P-value ; P ( X^2 > 3.01 ) = 0.99

since P-value > ∝ ( 0.05 ) then we will fail to reject H0 i.e. there is no change in the proportion of the viewing audience