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## E-Book Chapter Fifteen 15.18 Solutions

by: smartwriter Notetaker

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# E-Book Chapter Fifteen 15.18 Solutions

Marketplace > E Book Chapter Fifteen 15 18 Solutions
smartwriter Notetaker
CSU - Dominguez hills
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E-Book Chapter Fifteen 15.18 Solutions
COURSE
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TYPE
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PAGES
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WORDS
KARMA
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This 0 page Study Guide was uploaded by smartwriter Notetaker on Sunday November 15, 2015. The Study Guide belongs to a course at a university taught by a professor in Fall. Since its upload, it has received 22 views.

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### What is Karma?

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Date Created: 11/15/15
1513 Sixty four students in an ntmtileeterr eellege eeenemiee ellaee were aellredl haw entail1 e t he39ir had earned in eellege and ew eemtein they were abeut their eheiee of maj or R39eeettrelr margarine At a 2 1111 ie the degree efeerteinty independent e fquot eredite earned 3iquot Cer ta 111211r Credit Earned Very Uncertain Semewhet Efrain Few Ee ain Haw Tetal Q 12 E 3 23 159 8 11 1 22 er mere 1 T 1 1 19 Ce Intel 211 19 E4 E4 Credits Earned Very Uncertain Somewhat Certain Very Certain Row Total 0 9 12 8 3 23 10 59 8 4 10 22 60ormore171119 Col Total 21 19 24 64 12 8 3 4 10 1 7 11 We have here a chi square test for independence In the table above are the observed values In order to calculate the expected values we use a formula below Row total Column total Total The highlighted numbers in yellow are observed values and the numbers in bold are the totals for the specific row or column 2123 1923 2423 64 64 64 2122 1922 2422 64 64 64 2119 1919 2419 64 64 64 Above shown are the calculated for the expected values and they come out to be rounded to 3 decimal places 7547 6828 8625 7219 6531 825 6234 5641 7125 Now that we know the expected values we can get on With the hypothesis test Null Hypothesis Alternative Hypothesis earned ChiSquare test for Independence Ot001 Step 1 anotheses Step 2 Decision rule and identify the test H0 The degrees of certainty are independent of the credits earned The degrees of certainty are not independent of the credits DF degreesof freedomr 1c 13 13 1224 Critical value Chi 1014 1328 density U2quot Illquot Ii 1 in chiSquare Reject the null hypothesis if the test statistic is greater than 1328 falls in critical region and if the pValue is less than 001 otherwise fail to reject Step 3 calculate the test statistic This in English states the summation of the observed frequency minus the expected frequency all squared diVided by the expected frequency Observed frequencies 12 8 3 4 10 1 7 11 Expected Frequencies 7547 6828 8625 7219 6531 825 6234 5641 7125 Step 4 calculate the test statistic 12 75472 8 68282 3 236252 11 71252 0 7547 6828 8625 7125 Z 2627429310084494 4394411 210745614 Z 1476290728 x 14763 Step 5 Annlv the decision rule and communicate results The pValue for this test is P x2gt14763 4x 00052 density U2quot Illquot 0 1390 2390 chisquare Since the pValue 0052 is less than the significance level of 001 and the test statistic is greater than the critical value 14763gt1328 one can reject the null hypothesis 00052lt 001 14763gt 1328 One can reject the null and conclude that the degree of certainty and credits earned are not independent It makes sense that this is true since the more certain a person is With hisher major heshe Will do more classes of that certain fields

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