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 30%, CBS 27%, NBC 26%, and independents 17%. A sample of 300 homes two weeks after a Saturday night schedule revision yielded the following viewing audience data:
ABC 94 homes, CBS 71 homes, NBC 88 homes, and independents 47 homes.
1. Test with α = 0.05 to determine whether the viewing audience proportions changed.2. State the null and alternative hypotheses.
3. Find the value of the test statistic. (Round your answer to three decimal places.)4. Find the p-value. (Round your answer to four decimal places.)

Answer 1

Answer:

Part 1: Do not reject H0. There has not been a significant change in the viewing audience proportions.(The above conclusion holds as the p-value is greater than the significance level of 0.05)

Part 2:

H0: $p(ABC) = 0.30, p(CBS) = 0.27, p(NBC) = 0.26, p(IND) = 0.17$

Ha:  $p(ABC) \neq 0.30, p(CBS) \neq 0.27, p(NBC) \neq 0.26, p(IND) \neq 0.17$

Part 3: The test statistic is given as 3.0081.

Part 4: The p-value is given as  0.39037

Step-by-step explanation:

As per the data

Part 2

H0: $p(ABC) = 0.30, p(CBS) = 0.27, p(NBC) = 0.26, p(IND) = 0.17$

Ha:  $p(ABC) \neq 0.30, p(CBS) \neq 0.27, p(NBC) \neq 0.26, p(IND) \neq 0.17$

Part 3

As per the sample is of 300 homes, the expected values are given as

ABC=p(ABC)*300=0.30*300=90

CBS=p(CBS)*300=0.27*300=81

NBC=p(NBC)*300=0.26*300=78

IND=p(IND)*300=0.17*300=51

Now the values actually are given as

ABC=94

CBS=71

NBC=88

IND=47

So the test static is given as

$=\sum^{4}_{i=1}\dfrac{{O_i-E_i}^2}{E_i}\\=\dfrac{{O_1-E_1}^2}{E_1}+\dfrac{{O_2-E_2}^2}{E_2}+\dfrac{{O_3-E_3}^2}{E_3}+\dfrac{{O_4-E_4}^2}{E_4}\\=\dfrac{(94-90)^2}{90}+\dfrac{(71-81)^2}{81}+\dfrac{(88-78)^2}{78}+\dfrac{(47-51)^2}{51}\\=3.0081$

So the test statistic is given as 3.0081.

Part 4

Now the degree of freedoms are given as 4-1=3

and chi square  is given as 3.0081 so for α = 0.05 , the p-value is given as  0.39037

Part 1.

As the value of p in part 4 is greater than the value of α = 0.05, thus Do not reject H0. There has not been a significant change in the viewing audience proportions.(The above conclusion holds as the p-value is greater than the significance level of 0.05)