Module 8 - by performing a chi-square goodness-of fit test...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 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 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 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 v 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 scores l by the quare Test for Score / Data Set #1...
View Full Document

This note was uploaded on 08/04/2011 for the course ECN 601 taught by Professor Professor during the Spring '10 term at Grand Canyon.

Page1 / 44

Module 8 - by performing a chi-square goodness-of fit test...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online