1. 361 students recorded the number of hours of television they watch per week. The sample mean is 6.5 hours with a sample standard deviation 2.5 hours. Construct a 99% confidence interval for the population mean number of hours of television students watch per week. 2. In a survey of 1000 adults in the U.S., 20% say that they never exercise. Find a 90% confidence interval for the proportion of all adults in the U.S. who never exercise. 3. We have a sample of 2752 adults; 832 of them have children in the house and 1920 have no children in the house. They were asked if they visited a library in the last 12 months. 421 people who have children in the house visited a library and 810 people who do not have children in the house visited a library. Find a 92% confidence interval for pC − pN , the difference in proportions visiting the library between those with children in the house and those that have no children in the house. 4. We want to study the difference in means when a group of people pays individually for a dinner versus splitting the bill equally (as a group). We have a group of 40 people who split the bill equally – their sample mean cost was $50.92 with a sample standard deviation $14.33. We also have another group of 40 people who paid for dinners individually – their sample mean cost was $38.78 with a sample standard deviation of $12.54. Find a 95% confidence interval for μS − μI , the difference in population means costs between people who split their bills equally and people who pay their bills individually. 5. We would like to know what proportion of people would support taxes on junk food. If we have an estimate pˆ = 0.25 and if we want the margin of error to be ±3% (with 99% confidence), how many individuals should we sample?