Question

Advertisers fear that users of DVRs (digital video recorders) will “fast forward” past commercials when they...

Advertisers fear that users of DVRs (digital video recorders) will “fast forward” past commercials when they watch a recorded program. A leading British pay television company told its advertisers that this effect might be offset because DVR users watch more TV. A sample of 25 DVR users showed a daily mean screen time of 2 hours and 31 minutes with a standard deviation of 13 minutes, compared with a daily mean of 2 hours and 14 minutes with a standard deviation of 12 minutes for a sample of 25 non-DVR users.

(a-1)
Construct a 90 percent confidence interval for the difference in mean TV watching. (Round your answers to 2 decimal places.)

The 90% confidence interval is from  to

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
Advertisers fear that users of DVRs (digital video recorders) will “fast forward” past commercials when they...
Advertisers fear that users of DVRs (digital video recorders) will “fast forward” past commercials when they watch a recorded program. A leading British pay television company told its advertisers that this effect might be offset because DVR users watch more TV. A sample of 15 DVR users showed a daily mean screen time of 2 hours and 26 minutes with a standard deviation of 14 minutes, compared with a daily mean of 2 hours and 7 minutes with a standard...
In order to determine how many hours per week freshmen college students watch television, a random...
In order to determine how many hours per week freshmen college students watch television, a random sample of 25 students was selected. It was determined that the students in the sample spent an average of 19.5 hours with a sample standard deviation of 3.9 hours watching TV per week. Please answer the following questions: (a) Provide a 95% confidence interval estimate for the average number of hours that all college freshmen spend watching TV per week. (b) Assume that a...
A research council wants to estimate the mean length of time (in minutes) that the average...
A research council wants to estimate the mean length of time (in minutes) that the average U.S. adult spends watching television using digital video recorders (DVR’s) each day. To determine the estimate, the research council takes random samples of 35 U.S. adults and obtains the following times in minutes. 24 27 26 29 33 21 18 24 23 34 17 15 19 23 25 29 36 19 18 22 16 45 32 12 24 35 14 40 30 19 14...
1. A sample size of n = 70 is drawn from a population with proportion p...
1. A sample size of n = 70 is drawn from a population with proportion p = 0.32. Let p̂ be the sample proportion. Find p̂m and p̂s . Round the standard deviation to four decimal places. Find )28.0p̂P ( > . Find )40.0p̂P ( < 2.  On a certain television channel, 20% of the commercials are local advertisers. A sample of 150 commercials is selected. Would it be unusual for more than 26% of the commercials to be local advertisers?...
A research council wants to estimate the mean length of time (in minutes) that the average...
A research council wants to estimate the mean length of time (in minutes) that the average U.S. adult spends watching television using digital video recorders (DVR’s) each day. To determine the estimate, the research council takes random samples of 35 U.S. adults and obtains the following times in minutes. 24 27 26 29 33 21 18 24 23 34 17 15 19 23 25 29 36 19 18 22 16 45 32 12 24 35 14 40 30 19 14...
A research council wants to estimate the mean length of time (in minutes) that the average...
A research council wants to estimate the mean length of time (in minutes) that the average U.S. adult spends watching television using digital video recorders (DVR’s) each day. To determine the estimate, the research council takes random samples of 35 U.S. adults and obtains the following times in minutes. 24 27 26 29 33 21 18 24 23 34 17 15 19 23 25 29 36 19 18 22 16 45 32 12 24 35 14 40 30 19 14...
A market research firm has been contracted to determine television viewing habits. The firm selected a...
A market research firm has been contracted to determine television viewing habits. The firm selected a random sample of 16 children during a week and obtained a mean of 24 hours and a standard deviation of 7 hours. a) State what assumption you must make before you can construct a confidence interval. b) Construct a 90% confidence interval for the mean number of hours children spend watching television. Include a statement. c) The margin of error in part (a) was...
suppose that for adults the average amount of time spent watching tv per day is normally...
suppose that for adults the average amount of time spent watching tv per day is normally distribution with a mean of 42 minutes and a standard devation of 12 minutes. 1) what percent of adults spend less than 30 minutes watching tv per day? A) .04 B) .08 C) .11 D) .13 E) .16 2) What percent of adults more than 50 minutes watching tv per day? A) 0.18 B) 0.25 C) 0.45 D) 0.62 E) 0.75 3) what percent...
12. In order to estimate the mean amount of time computer users spend on the internet...
12. In order to estimate the mean amount of time computer users spend on the internet each​ month, how many computer users must be surveyed in order to be 90​% confident that your sample mean is within 15 minutes of the population​ mean? Assume that the standard deviation of the population of monthly time spent on the internet is 211 min. What is a major obstacle to getting a good estimate of the population​ mean? Use technology to find the...
Question 1 A researcher wishes to find a 90% confidence interval estimate for an unknown population...
Question 1 A researcher wishes to find a 90% confidence interval estimate for an unknown population mean using a sample of size 25. The population standard deviation is 7.2. The confidence factor z α 2for this est Group of answer choices 1.28 1.645 1.96 Question 2 A 90% confidence interval estimate for an unknown population mean μ is (25.81, 29.51). The length of this CI estimate is Group of answer choices 1.96 1.85 3.7 1.645 Question 3 Data below refers...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT