week 3 - team A - 5 step hypothesis - Sadaf part - and conclusion

Week 3 - team A - 5 step hypothesis - Sadaf part - and conclusion

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The alternate hypothesis is H 1 is not equal to 2.0 Step-2: The selected level of significance is α = 0.05 Step-3: The test statistic used for this problem is Chi Square : X 2 = [(f o – f e ) 2 /f e ] Bath Units fo fe fo-fe (fo-fe) 2 fo-fe) 2 /fe 1.5 2 2.25 -0.25 0.0625 0.0277 2.0 3 2.25 0.75 0.5625 0.25 2.5 3 2.25 0.75 0.5625 0.25 3.0 3 2.25 0.75 0.5625 0.25 X 2 = 0.7777 According to Appendix-E, the critical value for df = 3 is 7.815. Goodness of Fit Test observed expected O - E (O - E)² / E % of chi-sq 2 2.250 -0.250 0.028 3.57 3 2.250 0.750 0.250 32.14 3 2.250 0.750 0.250 32.14 3 2.250 0.750 0.250 32.14 11 9.000 2.000 0.778 100.00 Warning: sums should be equal. .78 chi-square 3 df .8548 p-value Step-4: Formulate the decision rule. The critical value identified by Appendix-E is 7.815, and so, we will reject Ho if it is greater (to the right) than this value and accept Ho if it is less than (to the left) the critical value. Step-5: Make a decision to reject or not reject the null hypothesis.
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This note was uploaded on 10/10/2010 for the course STATS Stats301 taught by Professor Regis during the Spring '10 term at DeVry Irvine.

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Week 3 - team A - 5 step hypothesis - Sadaf part - and conclusion

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