Question

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-1) Construct a 95 percent confidence interval for the difference in mean TV watching. (Round your answers to 2 decimal places.)
The 95% confidence interval is from _______to________ .
(a-2) Would this sample support the company’s claim (i.e., is zero within the confidence interval for the mean difference)?
Yes/No
(b) State any assumptions that are needed.
The variances are equal. The variances are unequal.

Answer 1

To construct a 95% confidence interval for the difference in mean TV watching, calculate the mean and standard deviation for each sample. Calculate the standard error of the difference in means. Find the margin of error and construct the confidence interval. The sample supports the company's claim and assumes equal variances.

Step-by-step explanation:

To construct a 95% confidence interval for the difference in mean TV watching between DVR users and non-DVR users:

1. Calculate the mean and standard deviation for each sample.
2. Calculate the standard error of the difference in means.
3. Find the margin of error by multiplying the standard error by the appropriate value from the t-distribution.
4. Construct the confidence interval by subtracting the margin of error from the difference in means and adding it to the difference in means.

Based on the sample data provided, the 95% confidence interval is from -12.23 to 46.23. Zero is within the confidence interval, so the sample supports the company's claim. The assumptions made for this analysis are that the variances are equal.