Solutions for 02402 exam 15. December 20111

More documents

Recommendations

Info

Medical drug SumA B C DNo side effects 5 3 10 2 20Light side effects 32 32 21 40 125Serious side effects 13 15 19 8 55Sum 50 50 50 50 200Question VII.1 (14) If the entire data material is used a 95% confidence intervalfor the proportion of serious side effects becomes? The large sample formula for aconfidence interval page 280 is used with x = 55 (observed number of serious sideeffects) and n = 200 (total number of observations): (55/200 = 0.275)55 √200 ± 1.96 0.275(1 − 0.275)/200which gives answer 2: 0.213 formed for which theχ 2 -value and number of degrees of freedom (f) is? This is an r × c table as in section10.4 (page 292-295) with c = 4 and r = 3. Hence the degrees of freedom is 3 · 2 = 6and only in answer 4 the right version of the χ 2 − statistic is given where the expectedvalues are put in the denominators - NOT the observed values as in answer 1. Theexpected number for e.g. the number of no side effects for drug B (observed o 12 = 3)would be:20 · 50e 12 = = 5200So: The correct answer is 4:χ 2 = (5−5)2 + (3−5)2 + (10−5)2 + (2−5)2 + (32−31.25)2 + (32−31.25)2 + (21−31.25)2 + (40−31.25)25 5 5 5 31.25 31.25 31.25 31.25+ (13−13.75)2 + (15−13.75)2 + (19−13.75)2 + (8−13.75)2 and f = 613.75 13.75 13.75 13.75Question VII.3 (16) Apart from the overall test, an investigation is wanted to testthe hypothesis that the proportion of serious side effects is significantly lower for drugD than for the others. Hence, the following hypothesis test is performed:H 0 : p D = p othersH 1 : p D for this hypothesis test are? We use the test for thedifference of two proportions (page 288 at the bottom) with x 1 = 8, n 1 = 50 , x 2 = 47and n 2 = 150 (and hence ˆp = 55/200 = 0.275 :Z =8− 4750 150√0.275(1 − 0.275)(150 + 1150 ) = −2.10And the P-value is found from Table 3 as P (Z ≤ −2.10)) or in R by pnorm(-2.10)(since it is a one-sided test only the left side is used). So the correct answer is 1:Z = −2.10 and P-value = 0.0188
Exercise VIIIFive different brands of tablets with the same active compound are compared withrespect to their solubility. For each brand four tablets were investigated. For eachtablet, percent solubility is measured after the tablet have been kept in 1000 ml deionizedwater for a while. The following data are found:BrandA B C D E39 36 48 60 30%solubility 43 37 46 65 3238 37 43 62 3935 35 41 67 31In the following table for a usual oneway analysis of variance, some of the results areshown. Further, the means for the brands are given.(It can be assumed that the data follows a normal distribution with the same variancein each group)The means of the five brands are:DF Sums of Mean FSquares SquareBrand x w 585.925 61.2465Error y 143.50 9.567Total z 2487.20Brand Number MeanA 4 38.75B 4 36.25C 4 44.50D 4 63.50E 4 33.00Question VIII.1 (17) The values for x, y, z and w are? As this is a one-way ANOVAas in chapter 12, section 2 with k = 5 and N = 20, we know that x = k − 1 = 5 − 1 = 4and y = N − k = 20 − 5 = 15 and z = N − 1 = 19 (page 362). We also know thatSS(Tr)=SST-SSE and hencew = SS(T r) = 2487.20 − 143.50 = 2343.7So the correct answer is 4:x = 4 , y = 15 , z = 19 and w = 2343.7Question VIII.2 (18) The result of the hypothesis test of no difference in mean solubilityfor the five brands is? (Both conclusion and argument must be correct) First9
Page 1 and 2: Solutions for 02402 exam 15. Decemb
Page 3 and 4: x=c(44.9,44.2,44.6,44.8,44,45.1)>t.
Page 5 and 6: On a 5% significance level only one
Page 7: counts the ones corresponding to th
Page 11 and 12: The result, which then is the round
Page 13 and 14: Machine AverageI II III IVDay shift
Page 15: Question XII.1 (28) The percentage

Solutions for 02402 exam 15. December 20111

Create successful ePaper yourself

Delete template?

Save as template?