Weinberger and Spotts compare the use of humor in television ads in the United States and the United Kingdom. Suppose that independent random samples of television ads are taken in the two countries. A random of 400 television ads in the United Kingdom reveals that 142 use humor, while a random sample of 500 television ads in the United States reveals that 122 use humor.
a.Set up the null and the alternative hypotheses needed to determine whether the proportion of ads using humor in the United Kingdom differs from the proportion of ads using in the United States.
b.Test the hypotheses you set up in part (a) by using the test statistic approach and by using 0.01 level of significance.
c.Set up the hypotheses needed to attempt to establish that the difference between the proportions of U.K and U.S ads using humor is more than 0.05. Test the hypotheses by using the P- value approach and by setting á equal to.10,.o5,.01,.001. how much evidence id there that the difference between the proportions exceeds.05?
Ho: proportions =
Ha: proportions ≠
Use Chi Square test.
http://math.hws.edu/javamath/ryan/ChiSquare.html
a. The null hypothesis (H0) in this case would be that the proportion of ads using humor in the United Kingdom is equal to the proportion of ads using humor in the United States. The alternative hypothesis (Ha) would be that the proportion of ads using humor in the United Kingdom differs from the proportion of ads using humor in the United States.
H0: Proportion of ads using humor in the United Kingdom = Proportion of ads using humor in the United States
Ha: Proportion of ads using humor in the United Kingdom ≠ Proportion of ads using humor in the United States
b. To test the hypotheses, we can use the z-test for comparing two proportions. The formula for the test statistic is:
z = (p1 - p2) / √(p̂ * (1 - p̂) * (1/n1 + 1/n2))
Where p1 and p2 are the sample proportions, n1 and n2 are the sample sizes, and p̂ is the pooled sample proportion.
First, calculate the sample proportions for each country:
p̂1 = 142/400 = 0.355
p̂2 = 122/500 = 0.244
Next, calculate the pooled sample proportion:
p̂ = (142 + 122) / (400 + 500) = 0.308
Now plug the values into the formula and calculate the test statistic:
z = (0.355 - 0.244) / √(0.308 * (1 - 0.308) * (1/400 + 1/500))
z = 3.017
Using a significance level of 0.01, we can compare the test statistic to the critical value associated with an alpha of 0.01. If the test statistic is greater than the critical value or less than the negative of the critical value, we reject the null hypothesis.
c. To test the hypothesis that the difference between the proportions of U.K. and U.S. ads using humor is more than 0.05, we can use the p-value approach. The null hypothesis (H0) in this case would be that the difference is not more than 0.05, and the alternative hypothesis (Ha) would be that the difference is more than 0.05.
To calculate the p-value, we need to find the probability of obtaining a test statistic as extreme as the one observed (or more extreme) assuming the null hypothesis is true.
We already calculated the test statistic in part (b), which was z = 3.017. Looking up the p-value associated with the test statistic in a standard normal distribution table, we can find the probability of getting a z-value of 3.017 or higher.
Now, for different significance levels (α), we can compare the calculated p-value to the significance level to determine the strength of evidence. If the p-value is less than α, we reject the null hypothesis. The smaller the p-value, the stronger the evidence against the null hypothesis.