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soln07

# soln07 - [ATP 7.5 12.5 20 32.5 62.5 155 320 16 rate 0.067...

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Unformatted text preview: [ATP] 7.5 12.5 20 32.5 62.5 155 320 16 rate 0.067 0.097 0.119 0.149 0.185 0.191 0.195 1/rate 1/[ATP] 14.925 0.133 10.309 0.080 8.403 0.050 6.711 0.031 5.405 0.016 5.236 0.006 5.128 0.003 1/rate (s/µM) 12 8 (1/rate) = 76.122(1/[ATP]) + 4.5406 4 0 0.00 0.05 0.10 0.15 1/[ATP] (µM-1) Problem 3 The process described in the problem statement is described by the binomial distribution. For each trial (rolling the die) there are two possible outcomes rolling a 6 or not rolling a 6. For an unweighted die the probability of rolling a 6 is p=1/6. Here you are asked to find the P ­value for the null hypothesis, H0: p≥1/6 for rolling one 6 during n rolls. The P ­value here is the probability that you roll 1 or less sixes in n rolls given that p≥1/6. This is a one sided hypothesis test. Since the process follows the binomial distribution you can calculate the exact answer using the binomial distribution. For the binomial distribution the probability distribution function is: !! !! = ! (1 − !)!!! ! The P ­value is the cumulative binomial distribution from 0 to 1: ! ! !! = ! !! ! !! !! ! (1 − !)!!! ! ! ! (1 − !)! = (1 − !)! !(1 − !)!!! = !"(1 − !)!!! !! 0! ! − 0 !! !1 = 1! ! − 1 !0 = ! ! ! = (1 − !)! + !"(1 − !)!!! ! !! To test the null hypothesis: H0 p>=1/6 set p=1/6 in the above equation: ! ! − !"#\$% = ! !! ! ! = (5/6)! + ! (5/6)!!! 6 Calculating the P value for n=10, 20, 30, and 40 gives: n P ­value 10 0.484516749 20 0.130420267 30 0.029489042 40 0.006123401 For sufficiently large sample sizes of a binomial processes you can use a test statistic based on the normal distribution approximation of the binomial distribution to test a hypothesis. This is covered in section 9 ­5 of Montgomery and Runger. In class a sample size of n≥5 was suggested as the minimum sample size to use the following test statistic. Montgomery and Runger recommend samples for which np≥5 and n(1 ­p)≥5. For the following solution we will assume that n≥5 is sufficient to use the test statistic although there are greater deviations from the true P ­value determined above for n=10 and 20 which do not meet the criteria of Montgomery and Runger. The test statistic is defined as: ! − !" !! = !"(1 − !) where X is the number of success in the sample (rolling a 6), n is the sample size, and p is the Bernoulli probability. Note that np is the average of the binomial distribution, and !"(1 − !) is the standard deviation of the binomial distribution. Assuming a value of p for the binomial process also sets the mean and standard deviation for the process, allowing us to use the Z ­test. For this problem: X=1 p=1/6 n=10, 20, 30, or 40 Calculating Z0 for each n: 1 − ! /6 !! = !(5/36) n Z0 10  ­0.5657 20  ­1.4000 30  ­1.9596 40  ­2.4042 The P ­value is the probability of having a sample at least as extreme as the one you have. Since this is a one sided test: P ­value=P(z≤Z0) The test statistic is for the normal distribution you can find these probabilities tabulated in the Cumulative Standard Normal Distribution table in Appendix A. Rounding Z0 to the nearest hundredth to avoid interpolation gives: n Z0 P ­value=P(z≤Z0) 10  ­0.57 0.284339 20  ­1.40 0.080757 30  ­1.96 0.024998 40  ­2.40 0.008198 ...
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