Commercials
Suppose that a commercial is run once on television, once on the radio, and once in a newspaper. The advertising agency believes that any potential consumer has a 20% chance of seeing the add on television, a 20% chance of hearing it on the radio, and a 20% chance of reading it in the newspaper. In a telephone survey of 800 consumers, the number claiming to have been exposed to the ad 0, 1, 2 or 3 times are as follows.
0 1 2 3
Observed number of people: 434 329 35 2
At the 1% significance level, test the null hypothesis that the number of times any consumer saw the add follows a binomial distribution with P(success) = .2
Hint: P(0) = (.8)(.8)(.8) = .512; P(1) = (3)(.2)(.8)(.8); P(2) = (3)(.2)(.2)(.8); P(3) = (.2)(.2)(.2)
*Show all work.
To test the null hypothesis, we will perform a chi-square goodness-of-fit test.
Step 1: State the null and alternative hypotheses:
Null hypothesis (H0): The number of times any consumer saw the ad follows a binomial distribution with P(success) = 0.2.
Alternative hypothesis (Ha): The number of times any consumer saw the ad does not follow a binomial distribution with P(success) = 0.2.
Step 2: Set the significance level (α):
The significance level is given as 1% or 0.01.
Step 3: Calculate the expected frequencies:
To calculate the expected frequencies, we need to multiply the total number of consumers (800) by the theoretical probabilities for 0, 1, 2, and 3 exposures.
The theoretical probabilities are: P(0) = 0.512, P(1) = 0.384, P(2) = 0.0768, P(3) = 0.01024.
Expected Frequency for 0 exposures = (0.512) * 800 = 409.6
Expected Frequency for 1 exposure = (0.384) * 800 = 307.2
Expected Frequency for 2 exposures = (0.0768) * 800 = 61.44
Expected Frequency for 3 exposures = (0.01024) * 800 = 8.192
Step 4: Calculate the chi-square test statistic:
The chi-square test statistic is calculated as:
χ^2 = Σ((O - E)^2 / E)
Where Σ represents the sum of all values, O is the observed frequency, and E is the expected frequency.
Using the observed and expected frequencies given, we can calculate the chi-square test statistic as follows:
χ^2 = ((434 - 409.6)^2 / 409.6) + ((329 - 307.2)^2 / 307.2) + ((35 - 61.44)^2 / 61.44) + ((2 - 8.192)^2 / 8.192)
Step 5: Determine the degrees of freedom:
The degrees of freedom for a goodness-of-fit test is (number of categories - 1).
In this case, since we have 4 categories (0, 1, 2, and 3), the degrees of freedom = 4 - 1 = 3.
Step 6: Determine the critical value:
The critical value is found in the chi-square distribution table using the degrees of freedom and the significance level.
For 3 degrees of freedom and a significance level of 0.01, the critical value is approximately 11.345.
Step 7: Compare the calculated chi-square test statistic with the critical value:
If the calculated chi-square test statistic exceeds the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Using the given observed and expected frequencies and calculating the chi-square test statistic as mentioned earlier, we get:
χ^2 ≈ 9.890
Since the calculated chi-square test statistic (9.890) is less than the critical value (11.345), we fail to reject the null hypothesis.
Therefore, at the 1% significance level, there is not enough evidence to suggest that the number of times any consumer saw the ad does not follow a binomial distribution with P(success) = 0.2.
To test the null hypothesis that the number of times any consumer saw the ad follows a binomial distribution with P(success) = 0.2, we can use the chi-square goodness-of-fit test.
Step 1: Set up the hypotheses.
- Null hypothesis (H0): The observed data follows a binomial distribution with P(success) = 0.2
- Alternative hypothesis (Ha): The observed data does not follow a binomial distribution with P(success) = 0.2
Step 2: Define the significance level.
Let's assume a significance level of α = 0.01.
Step 3: Calculate the expected frequencies.
Using the hint provided, we can calculate the expected probabilities for each category:
P(0) = (0.8)(0.8)(0.8) = 0.512
P(1) = (3)(0.2)(0.8)(0.8) = 0.384
P(2) = (3)(0.2)(0.2)(0.8) = 0.096
P(3) = (0.2)(0.2)(0.2) = 0.008
Then, we can calculate the expected frequencies for each category by multiplying the expected probabilities by the total sample size:
Expected frequency for 0: 0.512 * 800 = 409.6
Expected frequency for 1: 0.384 * 800 = 307.2
Expected frequency for 2: 0.096 * 800 = 76.8
Expected frequency for 3: 0.008 * 800 = 6.4
Step 4: Calculate the test statistic.
We can use the chi-square goodness-of-fit test statistic formula:
χ^2 = ∑((O - E)^2 / E)
Where O is the observed frequency and E is the expected frequency for each category.
Using the observed frequencies provided:
Observed frequency for 0: O(0) = 434
Observed frequency for 1: O(1) = 329
Observed frequency for 2: O(2) = 35
Observed frequency for 3: O(3) = 2
Calculating the test statistic:
χ^2 = ((434 - 409.6)^2 / 409.6) + ((329 - 307.2)^2 / 307.2) + ((35 - 76.8)^2 / 76.8) + ((2 - 6.4)^2 / 6.4)
Step 5: Determine the critical value.
Since we have 4 categories and our significance level is 0.01, we will use a chi-square distribution with 4 - 1 = 3 degrees of freedom.
Using a chi-square distribution table or a calculator, we can find the critical value for α = 0.01 and 3 degrees of freedom to be approximately 11.345.
Step 6: Compare the test statistic with the critical value.
If the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the calculated test statistic is χ^2 = 7.541.
Since 7.541 < 11.345, we fail to reject the null hypothesis.
Step 7: Interpret the result.
Based on the data and the test performed, there is not enough evidence to conclude that the observed data significantly deviates from a binomial distribution with P(success) = 0.2 at the 1% significance level.