Please only answer part (c) In an article in the Journal of Advertising , Weinbe
ID: 3205917 • Letter: P
Question
Please only answer part (c)
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 142 use humor, while a random sample of 500 television ads in the United States reveals that 124 use humor.
(a) 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.
H0: p1 ? p2 (Click to select)= 0 versus Ha: p1 ? p2 (Click to select)= 0.
(b) 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.)
z
(Click to select)RejectDo not Reject H0 at each value of ?; (Click to select)extremely strongsomenonestrongvery strong evidence.
(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 .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.)
z p-valueExplanation / Answer
a. Null hypothesis is that there is no difference in proportion of TV ads using humor in United Kingdom and United States. Therefore, H0:p1=p2.
The alternative hypothesis is there is difference in proportion of TV ads using humor in United Kingdom and United States. Therefore, H0:p1=/=p2.
b. Assume both the samples are independent ones, and sample size, n1, n1-x1, n2 and n2-x2 are greater than 5. The number of events is denoted by x. Use, 2-proportion Z test to compute th etest statistic and corresponding p values.
The pooled sample proportion, phatp=(x1+x2)/(n1+n2), where, x denotes number of events, and n denotes trials.
=(142+124)/(400+500)
=0.107
Z=(p1hat-p2hat)/[sqrt {phatp(1-phatp)} sqrt {1/n1+1/n2}], where, phat denotes sample proportion, phatp denote pooled sample proportion.
=(142/400-124/500)/[sqrt {0.107(1-0.107)} sqrt{1/142+1/124}]
=(0.355-0.248)/[sqrt {0.107(1-0.107)} sqrt{0.007+0.008}]
=3.50
p value: 0.000
Per rule, reject H0, if p value is less than given alpha. For all the given alpha values, the p value is less than them. Therefore, reject H0, and conclude that there is significant difference in proportion of TV ads using humor in United Kingdom and United States.
Since, for all given alpha, the null hypothesis is rejected, therefore, there is strong evidence.
c.
Null hypothesis is that the proportion of TV ads using humor in United Kingdom is atmost 50% that of United States. Therefore, H0:p1-p2<=0.5.
The alternative hypothesis is that the proportion of TV ads using humor in United Kingdom is 50% more than taht of United States. Therefore, H0:p1-p2>0.5.
Assume both the samples are independent ones, and sample size, n1, n1-x1, n2 and n2-x2 are greater than 5. The number of events is denoted by x. Use, 2-proportion Z test to compute the test statistic and corresponding p values.
Z=(p1hat-p2hat)-(p1-p2)/[sqrt {phatp(1-phatp)/n1+p2hat(1-p2hat)/n2}]
={(142/400-124/500)-0.5}/[sqrt {142/400(1-142/400)/400}+{124/500(1-124/500)/500}]
=-12.78
p value: 1.000
Per rule, reject H0, if p value is less than given alpha. For all the given alpha values, the p value is higher than them. Therefore, fail to reject H0, and conclude that there is not sufficient sample evidence to conclude that proportion of TV ads using humor in United Kingdom is 50% more than taht of United States.