Solutions for 02402 exam 15. December 20111
Solutions for 02402 exam 15. December 20111 Solutions for 02402 exam 15. December 20111
Question II.3 (5) We want to test the hypothesis σx 2 = σy 2 against the 2-sided alternative.If a significance level of 10% is used the test statistic and critical value for thistest are? We use the book section 9.3, pages 272-273:F = s2 ys 2 x= 0.2030.18 = 1.13The 2-sided 10% critical value is F 0.05 with (5, 5) degrees of freedom is 5.05 to be foundin R as qf(0.95,5,5) or in Table 6(a). Hence the correct answer is 3: Test statistic:1.13, critical value: 5.05.Question II.4 (6) Assume that the variance of the measurements on the outer diameteris σ x = 0.43. A new study is planned in which the outer diameter should bedetermined with an accuracy corresponding to a 95% confidence interval of size ± 0.1.What number of measurements, n is required? We use the sample size formula onpage 208 with α = 0.05, σ = 0.43 and E = 0.1, so the correct answer is 1: Around(1.96 · 0.43/0.1) 2 ≈ 71Exercise IIIA company has outsourced the manufacturing of a gasket for one of their valves to acompany in China. The gaskets are received in very large lots (with many thousandsof gaskets). The company controls a batch by sampling 200 gaskets at random fromthe lot, these are classified as defective or intact. A lot is accepted if there are at most2 defective item among the controlled ones.Question III.1 (7) What is the approximate probability of accepting a lot if the percentageof defectives is 0.4%?P (Accepting a lot) = P (X ≤ 2)where X follows a binomial with n = 200 and p = 0.004. Hence the correct answeris 3: P (X ≤ 2) = 95%. (Most likely in reality the sampling will be done WITHOUTreplacement - and hence the proper distribution would have been the Hypergeometricdistribution, BUT we are only told that the total amount of lots N out of which wesample the n = 200 is a large number - many thousands - NOT what it is - which wewould have needed to use the Hypergeometric distribution. And we are saved by thefact that in such cases where n ≤ N/10, the binomial is a good approximation of thehypergeometric (page 90-91).Question III.2 (8) Out of 2000 tested gaskets, a total of 14 was defective. The Chinesecompany guarantees that the error rate is not more than 0.4%, so the followinghypothesis test is performed:H 0 : p = 0.004H 1 : p > 0.0044
On a 5% significance level only one of the following options can be a critical value fora ”large sample” test of this hypothesis. Which one? We would use the test statisticgiven in page 284:14 − 2000 · 0.004Z = √ = 2.132000 · 0.004 · 0.996But actually we only need the one-sided critical value from the standard normal (Z)distribution to answer the question, and hence the correct answer is 4: z 0.05 = 1.645Question III.3 (9) We repeat the setup from the previous question: (hence one doesnot need to look at that again). Out of 2000 tested gaskets, a total of 14 was defective.The Chinese company guarantees that the error rate is not more than 0.4%, so thefollowing hypothesis test is performed:H 0 : p = 0.004H 1 : p > 0.004The P-value for this test can be found using the binomial distribution as follows? (Xis now the number of defective out of 2000). Instead of using the (large sample) Z-testas in the previous question, we are now asked to consider the (exact) binomial basedP-value probability computation. Since a P-value is the probability of observing whatwe’ve seen or something more extreme it becomes:P − value = P (X ≥ 14) = 1 − P (X ≤ 13)where X follows a binomial with n = 2000 and p = 0.004. In R this number canbe found directly as 1-pbinom(13,2000,0.004) to achieve 0.03388445 and hence thecorrect answer is 1: P (X ≥ 14) = 0.034. (Without using R, answer option 2 canbe excluded either by noting that the binomial in question has mean np = 8, so theprobability of going beyond 13 could NOT be as large as 0.999, OR in more detail toapply the normal approximation of the binomial that would say the same thing)Exercise IVThe arrival of guests wishing to check into a hotel is assumed in the period between 14(2pm) and 18 (6 pm) o’clock to be described by a poisson proces (arrivals are assumedevenly distributed over time and independent of each other). From extensive previousmeasurements it has been found that the probability that no guests arrive in a periodof 15 minutes is 0.30. (P (X = 0) = 0.30, where X describes the number of arrivals per15 min).Question IV.1 (10) The expected number of arrivals per 15 min, and the probabilitythat in a period of 1 hour 8 guests or more arrive are? First we need to find theintensity λ 15min of the poisson distribution for X - we use the general expression forgetting zero events in a poisson:P (X = 0) = λ0 15mine −λ 15min0!= e −λ 15min5
- Page 1 and 2: Solutions for 02402 exam 15. Decemb
- Page 3: x=c(44.9,44.2,44.6,44.8,44,45.1)>t.
- Page 7 and 8: counts the ones corresponding to th
- Page 9 and 10: Exercise VIIIFive different brands
- 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
Question II.3 (5) We want to test the hypothesis σx 2 = σy 2 against the 2-sided alternative.If a significance level of 10% is used the test statistic and critical value <strong>for</strong> thistest are? We use the book section 9.3, pages 272-273:F = s2 ys 2 x= 0.2030.18 = 1.13The 2-sided 10% critical value is F 0.05 with (5, 5) degrees of freedom is 5.05 to be foundin R as qf(0.95,5,5) or in Table 6(a). Hence the correct answer is 3: Test statistic:1.13, critical value: 5.05.Question II.4 (6) Assume that the variance of the measurements on the outer diameteris σ x = 0.43. A new study is planned in which the outer diameter should bedetermined with an accuracy corresponding to a 95% confidence interval of size ± 0.1.What number of measurements, n is required? We use the sample size <strong>for</strong>mula onpage 208 with α = 0.05, σ = 0.43 and E = 0.1, so the correct answer is 1: Around(1.96 · 0.43/0.1) 2 ≈ 71Exercise IIIA company has outsourced the manufacturing of a gasket <strong>for</strong> one of their valves to acompany in China. The gaskets are received in very large lots (with many thousandsof gaskets). The company controls a batch by sampling 200 gaskets at random fromthe lot, these are classified as defective or intact. A lot is accepted if there are at most2 defective item among the controlled ones.Question III.1 (7) What is the approximate probability of accepting a lot if the percentageof defectives is 0.4%?P (Accepting a lot) = P (X ≤ 2)where X follows a binomial with n = 200 and p = 0.004. Hence the correct answeris 3: P (X ≤ 2) = 95%. (Most likely in reality the sampling will be done WITHOUTreplacement - and hence the proper distribution would have been the Hypergeometricdistribution, BUT we are only told that the total amount of lots N out of which wesample the n = 200 is a large number - many thousands - NOT what it is - which wewould have needed to use the Hypergeometric distribution. And we are saved by thefact that in such cases where n ≤ N/10, the binomial is a good approximation of thehypergeometric (page 90-91).Question III.2 (8) Out of 2000 tested gaskets, a total of 14 was defective. The Chinesecompany guarantees that the error rate is not more than 0.4%, so the followinghypothesis test is per<strong>for</strong>med:H 0 : p = 0.004H 1 : p > 0.0044
Change language