Module 8 - a.,normality ,

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by performing a chi-square goodness-of fit test Scores on a Corporate Finance Midterm Exam In order to calculate the chi Square Statistic, we calculate the standardised values of the given s This is done by deducting the Sample Mean form the individual scores and dividing the residual Sample Standard Deviation  From the given data of scores, we compute Sample Mean = 79.39 Sample Standard Deviation  = 7.26 Work is shown below: I Combined "My Info Needed Section" w/ the Std. values Score Score in ascending order Std. values 72 62 -2.39 95 63 -2.26 83 65 -1.98 85 68 -1.57 65 69 -1.43 82 70 -1.29 84 70 -1.29 74 70 -1.29 74 70 -1.29 82 70 -1.29 81 71 -1.16 75 71 -1.16 78 71 -1.16 78 71 -1.16 72 72 -1.02 73 72 -1.02 75 72 -1.02 79 72 -1.02 72 73 -0.88 75 73 -0.88 77 73 -0.88 71 74 -0.74 78 74 -0.74 83 74 -0.74 84 74 -0.74 71 74 -0.74 81 74 -0.74 a.  Report and interpret the computed pvalue, normality 
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82 74 -0.74 79 75 -0.6 71 75 -0.6 73 75 -0.6 89 75 -0.6 74 75 -0.6 75 75 -0.6 93 76 -0.47 74 76 -0.47 88 76 -0.47 83 76 -0.47 90 77 -0.33 82 77 -0.33 79 77 -0.33 62 78 -0.19 73 78 -0.19 88 78 -0.19 76 78 -0.19 76 78 -0.19 76 78 -0.19 80 78 -0.19 84 79 -0.05 84 79 -0.05 91 79 -0.05 70 79 -0.05 76 79 -0.05 74 80 0.08 68 80 0.08 80 80 0.08 87 80 0.08 92 81 0.22 84 81 0.22 79 81 0.22 80 82 0.36 91 82 0.36 74 82 0.36 69 82 0.36 88 82 0.36 84 83 0.5 83 83 0.5 87 83 0.5 82 83 0.5 72 83 0.5 97 83 0.5 88 84 0.63
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70 84 0.63 83 84 0.63 92 84 0.63 94 84 0.63 63 84 0.63 83 84 0.63 81 84 0.63 84 85 0.77 78 85 0.77 84 86 0.91 91 87 1.05 78 87 1.05 71 87 1.05 77 88 1.19 78 88 1.19 70 88 1.19 86 88 1.19 79 89 1.32 70 90 1.46 75 91 1.6 80 91 1.6 75 91 1.6 77 92 1.74 74 92 1.74 87 93 1.87 78 94 2.01 85 95 2.15 70 97 2.42 Let us divide the 100 observations into 10 bins Bins Observed Counts (< -2.0) 2 (-2.0, -1.5) 2 (-1.5, -1.0) 14 (-1.0, -0.5) 16 (-0.5, 0.0) 19 (0.0, 0.5) 18 (0.5, 1.0) 11 (1.0, 1.5) 9 (1.5, 2.0) 6 (> 2.0) 3 100 2 4 6 8 10 12 Chi-Sq Bin Occupation
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Null Hypothesis is that the distribution is Normal Alternate Hypothesis is that it is not Normally distributed Bins Norm. Prob. Expected Counts Observed counts Obs-Exp (< -2.0) 0.02 2.3 2 -0.3 (-2.0, -1.5) 0.04 4.4 2 -2.4 (-1.5, -1.0) 0.09 9.2 14 4.8 (-1.0, -0.5) 0.15 15 16 1 (-0.5, 0.0) 0.19 19.1 19 -0.1 (0.0, 0.5) 0.19 19.1 18 -1.1 (0.5, 1.0) 0.15 15 11 -4 (1.0, 1.5) 0.09 9.2 9 -0.2 (1.5, 2.0) 0.04 4.4 6 1.6 (> 2.0) 0.02 2.3 3 0.7 100 100 The chi-square statistic is (obs-exp)^2/exp.values So calculated chi-square statistic is 5.84 as shown in yellow Since the data is divided into 10 bins and we have estimated two parameters, the calcuulated va  against chi-square distribution with 10-1-2 = 7 degrees of freedom For this distribution, the critical value for 0.05 level of significance is 14.07. Since 5.84 < 14.07,   Null Hypothesis that the data are normally distributed 0
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scores l by the
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quare Test for Score / Data Set #1
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Chi Value 0.04 1.31 2.5 0.07 0 0.06 1.07 0 0.58 0.21 5.84 alue may be tested  we don’t reject the
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and reporting interpreting the computed p-value. Is it still reasonable for the manufacturer to assume that the amount of detergent in these boxes is normally distributed? Information Gath First we need to test that whether the average weight of the boxes is more than 500 grams o For that we assume that the sample id coming from a Normal Population.
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