1. The problem statement, all variables and given/known data In June 2010, chemical analyses were made of 85 water samples (each of unit volume) taken from various parts of a city lake, and the measurements of chlorine content were recorded. During the next two winters, the use of road salt was substantially reduced in the catchment areas of the lake. In June 2012, 110 water samples were analyzed and their chlorine contents were recorded. Calculation of the mean and the standard deviation for the two sets of the data gives: Chlorine Content 2010 2012 Mean 18.3 17.8 Standard Deviation 1.2 1.8 Do the data provide strong evidence that there is a reduction of average chlorine level in the lake water in 2012 compared to the level in 2010? (a) Test with α = 0.05. (b) Suppose that a further study establishes that, in fact, the average chlorine level in the lake water didn’t change from 2010 to 2012. Referring back to part (a), did your analysis lead to a (i) Type I error, (ii) Type II error or (iii) correct decision? (c) Construct a 95% confidence interval for the difference of the population means. 3. The attempt at a solution a) let μ1 represent the 2010 mean and μ2 the 2012 X is the mean of the 2010 sample and Y is the mean of the 2012 sample since the sample sizes n1 = 85 > 30 and n2 = 110 > 30 we use the Z test H0: μ1-μ2=0 H1: μ1-μ2>0 we reject H0 if Z>zα = z.05 = 1.645 Z > 1.645 Z = [(X - Y) - 0]/[sqrt(s1^2/n1 + s2^2/n2)] = (18.3 - 17.8)/sqrt((1.2)^2/85 + (1.8)^2/110) = .5/.2154 = 2.32 Z = 2.32 > 1.645, so we reject H0 and say that the data provides evidence that there was a reduction in chlorine level b) type 1 error (rejection of H0 when H0 is true) c) large sample (n>30) so X-Y ± zα/2 sqrt(s1^2/n1 + s2^2/n2) 18.3 - 17.8 ± (1.96)sqrt((1.2)^2/85 + (1.8)^2/110) .5 ± (1.96)(.2154) .5 ± .422 .078 < μ1 - μ2 < .922 Am I doing this correctly?