SolutionHW6 - STAT500 HW#6_solutions Homework 6 Solutions...

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STAT500 HW#6_solutions Homework 6 Solutions 1. a. This poses an interesting question since we never discussed “intuitive” decisions prior to performing an analysis. So here is an intuitive thought process. Since one assumption is that the each population follows a normal distribution, then one can consider the application of 68-95-99 rule. That is, 68%, 95% and 99% of all observations fall within 1, 2, or 3 standard deviations respectively. Using 95% and the 2 SD rule, we can take the lowest mean, -0.1605 and highest mean, 0.2735, and build “loose” 95% confidence intervals around each [for ease, use sqrt(n) = sqrt(6) = 2]: -0.1605 +/- (2*0.1767/2) = - 0.3372 to 0.0162 0.2735 +/- (2*0.2492/2) = - 0.0243 to 0.5227 Notice these intervals do not overlap. Without an overlap we can deduce that there is a difference between these two means, and thus not all means are equal. b. Below is the output of the ANOVA. The p-value of 0.011, compared to alpha of 0.05, suggests that our sample provides statistical evidence that not all means are equal. This is consistent with our intuitive decision in part a. At least one device provides Ph readings that differ, on average, than the other three devices. One-way ANOVA: Ph versus Device Source DF SS MS F P Device 3 0.5837 0.1946 4.85 0.011 Error 20 0.8029 0.0401 Total 23 1.3866 S = 0.2004 R-Sq = 42.10% R-Sq(adj) = 33.41% c. The p-value of our F test is 0.011 d. The three conditions are: 1. Independent random samples (assumed by what was stated in the problem). 2. Equal variances (assumed since the test of equal variances in MINITAB produce a Bartlett’s p-value of 0.748 and Levene’s p-value of 0.926, both greater than alpha of 0.05), and 3. the normal distribution assumption is satisfied by the probability plot of the residuals (p-value of 0.550). e. Random sampling of a wide range of Ph values could result in one device receiving much more variable sample than the remaining devices. Having a greater variability in samples may cause this device to have more variability in its readings compared to another device(s) that receive samples that are more similar. This sampling bias could
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This note was uploaded on 10/08/2010 for the course STAT 500 at Pennsylvania State University, University Park.

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SolutionHW6 - STAT500 HW#6_solutions Homework 6 Solutions...

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