Please Help On this AP Statistics Problem!?!?!? Thanks
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 that the number of times any consumer saw the ad follows a binomial distribution with a success probability of P(success) = 0.2, we need to perform a hypothesis test.
Let's break down the steps to conduct this hypothesis 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: Determine the test statistic.
- In this case, we will use the chi-square test statistic to assess whether the observed data differs significantly from the expected binomial distribution.
- The chi-square test statistic is calculated using the formula: Χ² = ∑((Observed - Expected)² / Expected), where ∑ denotes the sum of the calculations across all categories.
Step 3: Set the significance level.
- The question already states that we should use a 1% significance level.
Step 4: Calculate the expected frequencies.
- To calculate the expected frequencies, we need to multiply the total sample size (800) by the corresponding probabilities for each category.
- Expected frequency for 0 times: 800 * P(0) = 800 * 0.512 = 409.6
- Expected frequency for 1 time: 800 * P(1) = 800 * 0.384 ≈ 307.2
- Expected frequency for 2 times: 800 * P(2) = 800 * 0.0768 ≈ 61.44
- Expected frequency for 3 times: 800 * P(3) = 800 * 0.0128 ≈ 10.24
Step 5: Calculate the chi-square test statistic.
- Using the formula Χ² = ∑((Observed - Expected)² / Expected), we can calculate the chi-square test statistic:
- For 0 times: ((434 - 409.6)² / 409.6) ≈ 2.955
- For 1 time: ((329 - 307.2)² / 307.2) ≈ 1.416
- For 2 times: ((35 - 61.44)² / 61.44) ≈ 6.113
- For 3 times: ((2 - 10.24)² / 10.24) ≈ 6.203
- Summing up these values: Χ² = 2.955 + 1.416 + 6.113 + 6.203 ≈ 16.687
Step 6: Determine the critical value.
- Since the question specifies a 1% significance level, we need to find the corresponding critical value for the chi-square distribution with degrees of freedom equal to the number of categories minus 1.
- In this case, we have 4 categories, so the degrees of freedom is 4 - 1 = 3.
- Looking up the critical value from a chi-square distribution table or using statistical software, we find that the critical value for a 1% (0.01) significance level and 3 degrees of freedom is approximately 11.345.
Step 7: Compare the test statistic to the critical value.
- Since the test statistic (16.687) is greater than the critical value (11.345), we have enough evidence to reject the null hypothesis.
Step 8: Interpret the result.
- Based on the test, we have evidence to suggest that the number of times any consumer saw the ad does not follow a binomial distribution with a success probability of P(success) = 0.2 at the 1% significance level.
In conclusion, the data suggests that the advertising agency's belief that any potential consumer has a 20% chance of seeing the ad on each media platform may not be accurate.