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Bus/Econ 302 Week 5 Notes

by: Collin Nordstrom

Bus/Econ 302 Week 5 Notes BUS 302

Marketplace > University of Mississippi > ECON > BUS 302 > Bus Econ 302 Week 5 Notes
Collin Nordstrom
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These are the in-class notes for Dr. McGinness' Statistics II. Contents cover Chapter 10: Hypothesis Testing with the Sample Proportion. The notes are individual examples that were not discussed th...
Statistics II
Ronnie McGinness
Class Notes




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This 3 page Class Notes was uploaded by Collin Nordstrom on Monday September 12, 2016. The Class Notes belongs to BUS 302 at University of Mississippi taught by Ronnie McGinness in Fall 2016. Since its upload, it has received 6 views. For similar materials see Statistics II in ECON at University of Mississippi.


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Date Created: 09/12/16
Chapter 10 Hypothesis Testing with the Sample Proportion -Also Content from 12 Ex: We want to estimate the proportion of Ole Miss Students who went to The Grove on Saturday. A random sample of 400 students finds that 312 students attended The Grove on Saturday. Confidence interval for the true proportion are: 80% and 97% pp(read below)= /n= 0.78 or 78%  Precise but has no accuracy pp ±Ζ∝ ∙ ^(1−p) n √ n pp(or p-hat) is used as an estimator to refer to a proportion of the sample of the entire group. Now, the only thing we are not given is Ζ ∝ confidence n interval. To find we must first find: = 1-0.8 (80% given) = .20 ∝ = 0.2 =0.1 which now make our equation p(Z 2 2 ???)= 0.1 We then find, in our Z-chart the closest number to .1.  If two numbers are equally close, take the average value. In this case we don’t need to. After we find .1 on the Z-chart we come to the confidence interval= 1.28 p(Z 1.28)=  0.1003 (number given in Z-Chart) .78 ±(1.28)∙78(1−.78)= .78 ±(1.20.02071 √ 400 =.78 margin of error)=2.65 ±.02651¿ = .78 +.02651=.80651 = .78 - .02651= .75349 Makes the final answer (75.4%, 80.7%) With an 80% confidence that the true value of P falls inside of the interval. Now we must find the same answer with a 97% confidence. ∝=1−.97=.03 ∝ =.15 2 p Z≤??? =.15  Find in Z-Chart and get -2.17 p Z≤−2.17=.15 ∝ p(1−p) pp±Z2 ∙√ n Standard error since we don’t have pp .78(1−.78) .78 ±(2.17)∙ 400 √ .78±2.17∙.02071 .78±.04494(Margin of Error) (.7351, .8249) (73.5%, 82.5%) 97% of the time true value will fall within this confidence interval


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