Solutions for 02402 exam 15. December 20111

The result, which then is the rounded (the R-function round is in the last line appliedto round to two decimal points) percentiles of the bootstrap distribution of differencesof means:0.5% 1% 2.5% 5% 25% 50% 75% 95% 97.5% 99% 99.5%-4.93 -4.64 -4.14 -3.79 -2.54 -1.67 -0.80 0.40 0.83 1.29 1.52The 99% confidence interval for the difference between the two means based on thisis? We use Section 9.5.2 in the R-note - telling us that we need to read off the 0.5%and 99.5% percentiles of the bootstrapped difference values. Hence the correct answeris 4: [−4.93, 1.52] (Note that if these computations are re-done in R, the results willnot be exactly the same due to simulation uncertainty)Question IX.3 (22) We want to test the following hypothesis on level α = 0.05 basedon the information in the previous question:H 0 : µ F − µ G = −4H 1 : µ F − µ G < −4What is the conclusion and the most correct argument? We use Section 9.6.4 in theR-note - doing bootstrap based hypothesis testing for the 2-sample situation. We cansee that less then 5% of the bootstrap difference values are below −4 (since the 5%percentile is −3.79. Hence answer 3 is the correct one: The hypothesis is rejected sincethe P-value is less than 5%Exercise XIt is known that the quotient between surface ares and volume has an influence on howfast a tablet is dissolved. To investigate this the following data was collected:%dissolved(y)Surface area/Volume(mm 2 /mm 3 ) (x)0.60 0.75 0.90 1.05 1.5033.00 36.25 38.75 44.50 63.50In R the following has been run:x=c(0.6,.75,0.9,1.05,1.5)y=c(33,36.25,38.75,44.50,63.5)summary(lm(y~x))with the following result:Call:11