In an article in the Journal of Advertising , Weinberger and Spotts compare the
ID: 3150668 • Letter: I
Question
In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor in television ads in the United States and in the United Kingdom. Suppose that independent random samples of television ads are taken in the two countries. A random sample of 400 television ads in the United Kingdom reveals that 143 use humor, while a random sample of 500 television ads in the United States reveals that 125 use humor.
Set up the null and alternative hypotheses needed to determine whether the proportion of ads using humor in the United Kingdom differs from the proportion of ads using humor in the United States.
Test the hypotheses you set up in part a by using critical values and by setting equal to .10, .05, .01, and .001. How much evidence is there that the proportions of U.K. and U.S. ads using humor are different? (Round the proportion values to 3 decimal places. Round your answer to 2 decimal places.)
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 .05 (five percentage points). Test these hypotheses by using a p-value and by setting a equal to .10, .05, .01, and .001. How much evidence is there that the difference between the proportions exceeds .05? (Round the proportion values to 3 decimal places. Round your z value to 2 decimal places and p-value to 4 decimal places.)
Calculate a 95 percent confidence interval for the difference between the proportion of U.K. ads using humor and the proportion of U.S. ads using humor. Interpret this interval. Can we be 95 percent confident that the proportion of U.K. ads using humor is greater than the proportion of U.S. ads using humor? (Round the proportion values to 3 decimal places. Round your answers to 4 decimal places.)
Set up the null and alternative hypotheses needed to determine whether the proportion of ads using humor in the United Kingdom differs from the proportion of ads using humor in the United States.
Explanation / Answer
A)
Formulating the hypotheses
Ho: p1 - p2 = 0
Ha: p1 - p2 =/= 0 [ANSWER]
*************************
b)
Here, we see that pdo = 0 , the hypothesized population proportion difference.
Getting p1^ and p2^,
p1^ = x1/n1 = 0.3575
p2 = x2/n2 = 0.25
Also, the standard error of the difference is
sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] = 0.030809647
Thus,
z = [p1 - p2 - pdo]/sd = 3.489166834 = 3.49 [ANSWER]
***********
Also, the p value is, as this is 2 tailed,
P = 0.000484529
As this P is less than 0.001, then we
REJECT HO at each value of alpha. EXTREMELY STROG evidence. [ANSWER]
***************************************************************************
c)
Formulating the hypotheses
Ho: p1 - p2 <= 0.05
Ha: p1 - p2 > 0.05
Here, we see that pdo = 0.05 , the hypothesized population proportion difference.
Getting p1^ and p2^,
p1^ = x1/n1 = 0.3575
p2 = x2/n2 = 0.25
Also, the standard error of the difference is
sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] = 0.030809647
Thus,
z = [p1 - p2 - pdo]/sd = 1.866298539 [ANSWER, Z]
Also, the p value is
P = 0.030999805 [ANSWER, P VALUE]
REJECT HO at each value alpha = 0.10 and 0.05. EXTREMELY STRONG EVIDENCE. [Depends on your convention in class how strong is 0.031.]
****************************************
d)
For the 95% confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.025
z(alpha/2) = 1.959963985
Margin of error = z(alpha/2)*sd = 0.060385799
lower bound = p1^ - p2^ - z(alpha/2) * sd = 0.047114201
upper bound = p1^ - p2^ + z(alpha/2) * sd = 0.167885799
Thus, the confidence interval is
( 0.047114201 , 0.167885799 ) [ANSWER]
YES, the entire interval is above 0.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.