AMU BUSN603 WK4 Assignment

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American Military University *

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603 001

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Statistics

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Apr 3, 2024

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docx

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4

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1 Cholesterol Test Student Name American Public University BUSN603: Quantitative Analysis Dr. Danielle Babb July 3, 2021
2 Cholesterol Test You are told that a random sample of 150 people from Manchester, New Hampshire, have been given cholesterol tests, and 60 of these people had levels over the "safe" count of 200. Using Excel, construct a 95% confidence interval for the population proportion of people in New Hampshire with cholesterol levels over 200. Note: Note that this question requires confidence interval for a proportion, the interval that is likely to capture the proportion of all population members' that satisfy a specified property. Do you remember the formula? If not, here it is: p estimate + - z-critical value X sqrt (p estimate (1-p estimate)/n) Also note that, for population proportion, we use z-multiple not the t-multiple. So, you can ask, where is the population standard deviation? You don't need it. Look at the formula. Solution Confidence Intervals for proportions Sample statistic number of successes (x) 60 sample size(n) 150 confidence level 0.95 proportion of successes (p') 0.4 proportion of failures (q') 0.6 critical value 1.96 margin of error 0.078 confidence interval lower limit 0.3216 upper limit 0.4784 We are 95% confident that the population proportion of people in Manchester, New Hampshire, with cholesterol over 200 is from 0.322 to 0.478 i. Finding the level of significance 1-95/100 =1-0.95= 0.05 Two-sided test hence 0.05 is divided by 2 0.05/2 =0.025 ii. Z value computation Finding the required probability = 1-0.025=0.975
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