this question is based on the assumption of large sample test. i am providing just numeric. please fill the blanks for which options not provided by yourself but i have explained each.
In an article in the Joumal of Advertising, Weinberger and Spotts compare the use of humor...
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...
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. (a) Set up the null...
In an article in the Joumal of Adverdising, Weinberger and Spotts compare the use of humor in television ads in the United States and in the United Kinadom. Suppose that independent random samples of teilevision ads are taken in the two countries. A random sample of 400 television ads in the United Kingdom reveals that 140 use humor, while a random sample of 500 television ads in the United Stanes reveals that 128se humor (a) Set up the null and...
In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor in television ads in the United States and the United Kingdom. They found that a substantially greater percentage of U.K. ads use humor. (a) Suppose that a random sample of 380 television ads in the United Kingdom reveals that 140 of these ads use humor. Find a point estimate of and a 95 percent confidence interval for the proportion of all U.K. television ads...
Questions 15-16: In an article, the authors compare the use of humor in TV ads in the United States and in the United Kingdom. A random sample of 100 TV ads in the United States reveals that 45 use humor, while a random sample of 120 TV ads in the United Kingdom reveals that 35 use humor. Let pi and p2 represent the true proportion of ads using humor in the United States and in the United Kingdom respectively. If...
Suppose a group of marketing students have decided to compare the use of humor in television ads in the United Kingdom and the United States as part of a group project. Suppose they find that in a random sample of 400 United Kingdom television ads 142 use humor. In a sample of 500 television ads in the United States, 122 use humor. Test at the 0.05 signficiance level to see if there is a difference in the population proportion of...
Suppose a sample of 49 paired differences that have been randomly selected from a normally distributed population of paired differences yields a sample mean d¯ =5.0d¯ =5.0 of and a sample standard deviation of sd = 7.8. (a) Calculate a 95 percent confidence interval for µd = µ1 – µ2. Can we be 95 percent confident that the difference between µ1 and µ2 is greater than 0? (Round your answers to 2 decimal places.) Confidence interval = [ , ] ;...
The manufacturer of the ColorSmart-5000 television set claims that 95 percent of its sets last at least five years without needing a single repair. In order to test this claim, a consumer group randomly selects 396 consumers who have owned a ColorSmart-5000 television set for five years. Of these 396 consumers, 311 say that their ColorSmart-5000 television sets did not need repair, while 85 say that their ColorSmart-5000 television sets did need at least one repair. (a) Letting p be...
Suppose a sample of 49 paired differences that have been randomly selected from a normally distributed population of paired differences yields a sample mean of d? 5 and a sample standard deviation of sd-7. (a) Calculate a 95 percent confidence interval for μο" μι-P2. (Round your answers to 2 decimal places.) Confidence interval 2.989 7011 (b) Test the null hypothesis Ho: μ.-0 versus the alternative hypothesis Ha: Ha ? 0 by setting ? equal to 10, 05, 01, and .001....
Exercise 10.16 METHODS AND APPLICATIONS Suppose a sample of 49 paired differences that have been randomly selected from a normally distributed population of paired differences yields a sample mean of d = 5.9 and a sample standard deviation of sd = 6.9. (a) Calculate a 95 percent confidence interval for Hd = 41 - 42. Can we be 95 percent confident that the difference between H1 and P2 is greater than 0? (Round your answers to 2 decimal places.) Confidence...