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 deviation of 12 minutes for a sample of 15 non-DVR users. (a) Construct a 95 percent confidence interval for the difference in mean TV watching. Would this sample support the company’s claim (i.e., is zero within the confidence interval for the mean difference)? (b) Discuss any assumptions that are needed.
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 deviation of 12 minutes for a sample of 15 non-DVR users. (a) Construct a 95 percent confidence interval for the difference in mean TV watching. Would this sample support the company’s claim (i.e., is zero within the confidence interval for the mean difference)? (b) Discuss any assumptions that are needed.
Solution Summary: The author states that the sample does not support the company's claim.
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 deviation of 12 minutes for a sample of 15 non-DVR users. (a) Construct a 95 percent confidence interval for the difference in mean TV watching. Would this sample support the company’s claim (i.e., is zero within the confidence interval for the mean difference)? (b) Discuss any assumptions that are needed.
Definition Definition Number of subjects or observations included in a study. A large sample size typically provides more reliable results and better representation of the population. As sample size and width of confidence interval are inversely related, if the sample size is increased, the width of the confidence interval decreases.
Please could you explain why 0.5 was added to each upper limpit of the intervals.Thanks
28. (a) Under what conditions do we say that two random variables X and Y are
independent?
(b) Demonstrate that if X and Y are independent, then it follows that E(XY) =
E(X)E(Y);
(e) Show by a counter example that the converse of (ii) is not necessarily true.
1. Let X and Y be random variables and suppose that A = F. Prove that
Z XI(A)+YI(A) is a random variable.
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.
Hypothesis Testing using Confidence Interval Approach; Author: BUM2413 Applied Statistics UMP;https://www.youtube.com/watch?v=Hq1l3e9pLyY;License: Standard YouTube License, CC-BY
Hypothesis Testing - Difference of Two Means - Student's -Distribution & Normal Distribution; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=UcZwyzwWU7o;License: Standard Youtube License