Question

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 30%, CBS 28%, NBC 22%, and independents 20%. A sample of 300 homes two weeks after a Saturday night schedule revision yielded the following viewing audience data: ABC 93 homes, CBS 70 homes, NBC 82 homes, and independents 55 homes. Test with =.05 to determine whether the viewing audience proportions changed. Round your answers to two decimal places.

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: The Audience for sample of 300 homes is same as during the first 13 weeks.

Alternative hypothesis: At least one of the proportions in the null hypothesis is false.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.

Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = k - 1 = 4 - 1
D.F = 3
(E_i) = n * p_i

X² = 6.73

where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, E_i is the expected frequency count for level i, O_i is the observed frequency count for level i, and X² is the chi-square test statistic.

The P-value is the probability that a chi-square statistic having 3 degrees of freedom is more extreme than 6.73.

We use the Chi-Square Distribution Calculator to find P(X² > 6.73) = 0.665

Interpret results. Since the P-value (0.665) is greater than the significance level (0.05), we have to accept the null hypothesis.