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Mm207

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Submitted By bragg
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MM207 Final Project
175 total students took the survey
42+4+3+7+1+1 =58 students travel less than 5 miles to work
58/175=0.33142857142857142857142857142857
1-0.33142857142857142857142857142857=0.66857142857142857142857142857143
(1.96/.02)^2*0.33142857142857142857142857142857*0.66857142857142857142857142857143
=9604 * 0.33142857142857142857142857142857 = 3183.04
=3183.04 * 0.66857142857142857142857142857143
=2128.0896
173 answered the question
42+4+3+7+1+1 =58 students travel less than 5 miles to work
58/173=0.3352601156069364161849710982659
1-0.3352601156069364161849710982659=0.6647398843930635838150289017341
(1.96/.02)^2 * 0.6647398843930635838150289017341 =
= 9604*0.3352601156069364161849710982659*0.6647398843930635838150289017341
=2140.3548397874970764141802265361
10. (-1) A professor at Kaplan University claims that the average age of all Kaplan students is 36 years old. Use a 95% confidence interval to test the professor's claim. Is the professor's claim reasonable or not? Explain.
My answer:
YES, it is because the intervals are roughly 36-419the professor’s claim is pretty accurate. But, 36<36.36. This is where Statistics draws the hard line, no more “roughly” or
“approximately”. If the Prof had said 36.25, you would still reject because 36.25<36.36. interval are 36.36-41.137 size= 175 mean=37.94857 standard deviation=10.726628 sum=6641 6641/175=37.948571428571428571428571428571 or 37.95 sqrt of 175=13.228756555322952952508078768196 st deviation/sqrt of 175
= 10.726628/13.228756555322952952508078768196=0.81085685983720420676031901680653 z = 1.96 or 95% MM207 Final Project margin of error = 1.96 * 0.81085685983720420676031901680653
= 1.5892794452809202452502252729408
37.94857-1.5892794452809202452502252729408=36.359290554719079754749774727059
37.94857+1.5892794452809202452502252729408=41.127128890561840490500450545882

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