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.)

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:

Ha:  

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:

Ha:  

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

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)

answer: what do you mean