Question

In an article in the Journal 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 the United Kingdom. They found that a substantially greater percentage of U.K. ads use humor.

(a) Suppose that a random sample of 375 television ads in the United Kingdom reveals that 141 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 that use humor. (Round your answers to 3 decimal places.)   

pˆp^ =
The 95 percent confidence interval is [,].

(b) Suppose a random sample of 533 television ads in the United States reveals that 117 of these ads use humor. Find a point estimate of and a 95 percent confidence interval for the proportion of all U.S. television ads that use humor. (Round your answers to 3 decimal places.)

pˆp^ =
The 95 percent confidence interval is [,].


(c) Do the confidence intervals you computed in parts a and b suggest that a greater percentage of U.K. ads use humor?


(No or Yes) the U.K. 95 percent confidence interval is (not above or above) the maximum value
in the confidence interval for the U.S.

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
In an article in the Journal 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 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...
In an article in the Journal 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 141 use humor, while a random sample of 500 television ads in the United States reveals that 122 use humor. (a) Set up the null...
In an article in Marketing Science, Silk and Berndt investigate the output of advertising agencies. They...
In an article in Marketing Science, Silk and Berndt investigate the output of advertising agencies. They describe ad agency output by finding the shares of dollar billing volume coming from various media categories such as network television, spot television, newspapers, radio, and so forth. (a) Suppose that a random sample of 400 U.S. advertising agencies gives an average percentage share of billing volume from network television equal to 7.46 percent, and assume that σ equals 1.41 percent. Calculate a 95...
Show all your work. Indicate clearly the methods you use, because you will be scored on...
Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. A recent survey collected information on television viewing habits from a random sample of 1,000 people in the United States. Of those sampled, 37 percent indicated that their favorite sport to watch on television was American football. (a) Construct and interpret a 95 percent confidence interval...
Use the normal distribution to find a confidence interval for a proportion p given the relevant...
Use the normal distribution to find a confidence interval for a proportion p given the relevant sample results. Give the best point estimate for p, the margin of error, and the confidence interval. Assume the results come from a random sample. A 95% confidence interval for the proportion who will answer ‘‘Yes” to a question, given that 60 answered yes in a random sample of 95 people. Round your answers to three decimal places. Point estimate = Margin of error...
A newspaper article summarized data from a survey of 1,855 recruiters and human resource professionals. The...
A newspaper article summarized data from a survey of 1,855 recruiters and human resource professionals. The article indicated that 51% of the people surveyed had reconsidered a job candidate based on his or her social media profile. Assume that the sample is representative of the population of recruiters and human resource professionals in the United States. (a) Use the given information to estimate the proportion of recruiters and human resource professionals who have reconsidered a job candidate based on his...
1. A poll of 2234 U.S. adults found that 31% regularly used Facebook as a news...
1. A poll of 2234 U.S. adults found that 31% regularly used Facebook as a news source. Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 90% level of confidence. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval: % to % Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use...
Chapter 6, Section 1-CI, Exercise 011 Use the normal distribution to find a confidence interval for...
Chapter 6, Section 1-CI, Exercise 011 Use the normal distribution to find a confidence interval for a proportion p given the relevant sample results. Give the best point estimate for p, the margin of error, and the confidence interval. Assume the results come from a random sample. A 95% confidence interval for p given that p^=0.34 and n=475. Round your answer for the best point estimate to two decimal places, and your answers for the margin of error and the...
The 2003 Statistical Abstract of the United States reported the percentage of people 18 years of...
The 2003 Statistical Abstract of the United States reported the percentage of people 18 years of age and older who smoke. Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of 30. a. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of .02? Use 95% confidence. Remove all commas from your answer before submitting....
Chapter 6, Section 1-CI, Exercise 012 Use the normal distribution to find a confidence interval for...
Chapter 6, Section 1-CI, Exercise 012 Use the normal distribution to find a confidence interval for a proportion given the relevant sample results. Give the best point estimate for , the margin of error, and the confidence interval. Assume the results come from a random sample. A 95% confidence interval for p given that p-hat = 0.9 and n = 120 . Round your answer for the point estimate to two decimal places, and your answers for the margin of...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT