In Los Angeles there are three network television stations, each with it own evening news program from 6:00 to 6:30 PM. According to a report in this morning’s local newspaper, a random sample of 150 viewers last night revealed 53 watched the news on KNBC (channel 2), 64 watched KABC (channel 7), and 33 viewed KCBS (channel 2). At the 0.05 significance level, is there a difference in the proportion of viewers watching the three channels?
State the null and alternative hypothesis.
. Identify the level of significance: Given at 0.05
3. Calculate the test statistic
4. Formulate the decision rule.
5. Make a decision.
1. The null and alternative hypotheses can be stated as follows:
- Null hypothesis (H0): There is no difference in the proportion of viewers watching the three channels.
- Alternative hypothesis (Ha): There is a difference in the proportion of viewers watching the three channels.
2. The level of significance is given at 0.05. This means that we will reject the null hypothesis if the probability of obtaining our sample data (or more extreme) under the assumption of the null hypothesis is less than 0.05.
3. To test the hypothesis, we need to calculate the test statistic. In this case, we will use the chi-square test for independence since we are comparing proportions across different categories. The formula for the test statistic is:
χ² = Σ [(Oij - Eij)² / Eij]
Where:
- Oij represents the observed frequency in each cell (i, j).
- Eij represents the expected frequency in each cell (i, j) under the assumption of independence.
We first need to calculate the expected frequencies. To do this, we assume that the proportions are the same for all three channels. We can calculate the expected frequency for each cell by multiplying the total sample size by the assumed proportions.
Expected frequency for KNBC = (53+33) * (150/150) = 86
Expected frequency for KABC = 64 * (150/150) = 64
Expected frequency for KCBS = (53+33) * (150/150) = 86
Now, we use the formula to calculate the chi-square test statistic:
χ² = [(53-86)² / 86] + [(64-64)² / 64] + [(33-86)² / 86]
4. To formulate the decision rule, we need to determine the critical value of chi-square at the given level of significance (0.05) and degrees of freedom. The degrees of freedom is calculated as (number of rows - 1) * (number of columns - 1). In this case, we have (3-1) * (3-1) = 2 * 2 = 4.
We look up the critical value of chi-square with 4 degrees of freedom and a significance level of 0.05 from the chi-square distribution table. Let's assume the critical value is 9.488 (hypothetical value).
The decision rule is:
- If the test statistic is greater than the critical value (9.488), we reject the null hypothesis.
- If the test statistic is less than or equal to the critical value (9.488), we fail to reject the null hypothesis.
5. Make a decision:
- Calculate the test statistic using the formula mentioned in step 3.
- Compare the test statistic with the critical value obtained in step 4.
- If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that there is a difference in the proportion of viewers watching the three channels.
- If the test statistic is less than or equal to the critical value, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest a difference in the proportion of viewers watching the three channels.