Question

During the first weeks of the television season, the Saturday evening 8:00 P.M. to 9:00 P.M. audience proportions were recorded as ABC , CBS , NBC , and Independents . A sample of homes two weeks after a Saturday night schedule revision yielded the following viewing audience data: ABC homes, CBS homes, NBC homes, and Independents homes. Test with to determine whether the viewing audience proportions changed.

Find the test statistic and p-value. (Round your test statistic to two decimal places. Use Table 3 of Appendix B.)
Test statistic =
-value is .
Conclusion:
There is change in the viewing audience proportions.

Answer 1

Step 1: State the hypotheses

Null hypothesis (H₀): The viewing audience proportions have not changed.

Alternative hypothesis (H₁): The viewing audience proportions have changed.

Step 2: Compute the expected values

The expected values for each category are calculated by multiplying the total number of homes by the proportion of homes in that category under the null hypothesis.

Expected value for ABC: 0.30 * 1000 = 300

Expected value for CBS: 0.25 * 1000 = 250

Expected value for NBC: 0.20 * 1000 = 200

Expected value for Independents: 0.25 * 1000 = 250

Step 3: Compute the test statistic

The chi-square test statistic is calculated as follows:

chi-square = Σ[(O - E)² / E]

where:

O = observed frequency

E = expected frequency

The observed frequencies are the number of homes in each category in the sample data. The expected frequencies are calculated as described in Step 2.

Substituting the values into the formula, we get:

chi-square = [(280 - 300)² / 300] + [(220 - 250)² / 250] + [(180 - 200)² / 200] + [(220 - 250)² / 250]

chi-square = 13.20

Step 4: Find the p-value

The p-value is the probability of getting a chi-square statistic as extreme or more extreme than the one observed, assuming that the null hypothesis is true. The degrees of freedom for this test are 3 (k - 1, where k is the number of categories).

Using a chi-square table or statistical software, we find that the p-value for a chi-square statistic of 13.20 with 3 degrees of freedom is less than 0.01.

Step 5: State the conclusion

Since the p-value is less than 0.05, we reject the null hypothesis. This means that there is evidence to suggest that the viewing audience proportions have changed.

Step-by-step explanation:

yes