Unit 3 Extra Credit: Problem 3 (1 point) Periodically, the county Water Department tests the drinking water of homeowners for contaminants such as lead and copper. The lead and copper levels in water specimens collected in 1998 for a sample of 10 residents of a subdevelopement of the county are shown below. lead (\"g/L) copper (mg/L) 1 0.487 2.6 0.642 3.2 0.871 0.6 0.017 4.3 0.58 1.6 0.178 1.1 0.278 5.3 0.2 5.7 0.863 0.8 0.839 (a) Construct a 99% confidence interval for the mean lead level in water specimans of the subdevelopment. Sits (b) Construct a 99% confidence interval for the mean copper level in water specimans of the subdevelopment. S It 5 Unit 3 Extra Credit: Problem 4 (1 point) In cases where the confidence level leads to an area underneath the probability density function for either the standard normal or the t distribution that is not exactly listed on the tables, linear interpolation is used. We have seen a little bit of this in the 2,, for the 90% and 99% confidence levels. At 99% confidence, the significance level, a, is 0.01 and the area underneath the standard normal pdf to the left of Zn, is 0.995. The area 0.995 is halfway in between 0.9949 and 0.9951 on the ztable, therefore 2,, is 2.575, the value halfway between 2.57 and 2.58. Similarly, at 91% confidence, the area underneath the standard normal pdf to the left of z\" is 0.955. The value, 0.955, is about 0.555555 of the distance between 0.9545 and 0.9554, the two closest values found on the ztable, and therefore 2,, should be the value about 0.555555 of the distance between 1.69 and 1.70, or 1.695555. Try this out on the following problem. A medical statistician wants to estimate the average weight loss of people who are on a new diet plan. Assume that the standard deviation of the population of weight loss is about 4.5 pounds. In a sample of 19 subjects on the new diet plan, the average average weight loss was 10 pounds. What is the 98% confidence inten/al estimate for the average weight loss for anyone that follows the new diet plan? 514$