Chapter 9: Problem 476
Given the following set of ungrouped measurements $$ 3,5,6,6,7, \text { and } 9 $$ determine the mean, median, and mode.
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Use the Kolmogorov-Smirnov Statistic to find a \(95 \%\) confidence interval for \(\mathrm{F}(\mathrm{x}) . \mathrm{F}(\mathrm{x})\) is the cumulative distribution function of a population from which the following ordered samples was taken: \(8.2,10.4,10.6,11.5,12.6,12.9\), \(13.3,13.3,13.4,13.4,13.6,13.8,14.0,14.0,14.1,14.2\) \(14.6,14.7,14.9,15.0,15.4,15.6,15.9,16.0,16.2,16.3\) 17.2,17.4,17.7,18.1 .$
Flapjack Computers is interested in developing a new tape drive for a proposed new computer. Flapjack does not have research personnel available to develop the new drive itself and so is going to subcontract the development to an independent research firm. Flapjack has set a fee of \(\$ 250,000\) for developing the new tape drive and has asked for bids from various research firms. The bid is to be awarded not on the basis of price (set at \(\$ 250,000\) ) but on the basis of both the technical plan shown in the bid and the firm's reputation. Dyna Research Institute is considering submitting a proposal (i.e., a bid) to Flapjack to develop the new tape drive. Dyna Research Management estimated that it would cost about \(\$ 50,000\) to prepare a proposal; further they estimated that the chances were about \(50-50\) that they would be awarded the contract. There was a major concern among Dyna Research engineers concerning exactly how they would develop the tape drive if awarded the contract. There were three alternative approaches that could be tried. One involved the use of certain electronic components. The engineers estimated that it would cost only \(\$ 50,000\) to develop a prototype of the tape drive using the electronic approach, but that there was only a 50 percent chance that the prototype would be satisfactory. A second approach involved the use of certain magnetic apparatus. The cost of developing a prototype using this approach would cost \(\$ 80,000\) with 70 percent chance of success. Finally, there was a mechanical approach with cost of \(\$ 120,000\), but the engineers were certain of success. Dyna Research could have sufficient time to try only two approaches. Thus, if either the magnetic or the electronic approach tried and failed, the second attempt would have to use the mechanical approach in order to guarantee a successful prototype. The management of Dyna Research was uncertain how to take all this information into account in making the immediate decision-whether to spend \(\$ 50,000\) to develop a proposal for Flapjack. Can you help?
Suppose we want to compare 2 treatments for curing acne (pimples). Suppose, too, that for practical reasons we are obliged to use a presenting sample of patients. We might then decide to alternate the 2 treatments strictly according to the order in which the patients arrive \((\mathrm{A}, \mathrm{B}, \mathrm{A}, \mathrm{B}\), and so on). Let us agree to measure the cure in terms of weeks to reach \(90 \%\) improvement (this may prove more satisfactory than awaiting \(100 \%\) cure, for some patients, may not be completely cured by either treatment, and many patients might not report back for review when they are completely cured). This design would ordinarily call for Wilcoxon's Sum of Ranks Test, but there is one more thing to be considered: severity of the disease. For it could happen that a disproportionate number of mild cases might end up, purely by chance, in one of the treatment groups, which could bias the results in favor of this group, even if there was no difference between the 2 treatments. It would clearly be better to compare the 2 treatments on comparable cases, and this can be done by stratifying the samples. Suppose we decide to group all patients into one or other of 4 categories \- mild, moderate, severe, and very severe. Then all the mild cases would be given the 2 treatments alternatively and likewise with the other groups. Given the results tabulated below (in order of size, not of their actual occurrence), is the evidence sufficient to say that one treatment is better than the other? $$ \begin{array}{|c|c|c|} \hline \text { Category } & \begin{array}{c} \text { Treatment A } \\ \text { Weeks } \end{array} & \begin{array}{c} \text { Treatment B } \\ \text { Weeks } \end{array} \\ \hline \text { (I) Mild } & 2 & 2 \\ & 3 & 4 \\ \hline \text { (II) Moderate } & 3 & 4 \\ & 5 & 6 \\ & 6 & 7 \\ & 10 & 9 \\ \hline \text { (III) Severe } & 6 & 9 \\ & 8 & 14 \\ & 11 & 14 \\ \hline \text { (IV) Very severe } & 8 & 12 \\ & 10 & 14 \\ & 11 & 15 \\ \hline \end{array} $$
Let \(\mathrm{T}\) be distributed with density function \(f(t)=\lambda e^{-\lambda . t} \quad\) for \(t>0\) and \(=0\) otherwise If \(S\) is a new random variable defined as \(S=\) In \(\mathrm{T}\), find the density function of \(\mathrm{S}\).
Consider the joint distribution of \(\mathrm{X}\) and \(\mathrm{Y}\) given in the form of a table below. The cell (i,j) corresponds to the joint probability that \(\mathrm{X}=\mathrm{i}, \mathrm{Y}=\mathrm{j}\), for \(\mathrm{i}=1,2,3, \mathrm{j}=1,2,3\) $$ \begin{array}{|c|c|c|c|} \hline \mathrm{Y}^{\mathrm{X}} & 1 & 2 & 3 \\ \hline 1 & 0 & 1 / 6 & 1 / 6 \\ \hline 2 & 1 / 6 & 0 & 1 / 6 \\ \hline 3 & 1 / 6 & 1 / 6 & 0 \\ \hline \end{array} $$ Check that this is a proper probability distribution. What is the marginal distribution of \(\mathrm{X} ?\) What is the marginal distribution of \(\mathrm{Y}\) ?
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