1.Build a 99% confidence interval for the percentage of black beans in Class1? A
ID: 3229859 • Letter: 1
Question
1.Build a 99% confidence interval for the percentage of black beans in Class1? And Class 2?
Population1_Class 1
Population2_Class 2
Student# Black Beans While Beans Total % black beans 1 18 16 34 0.529 2 15 16 31 0.484 3 15 15 30 0.500 4 15 18 33 0.455 5 15 17 32 0.469 6 17 14 31 0.548 7 15 17 32 0.469 8 21 19 40 0.525 9 13 22 35 0.371 10 18 16 34 0.529 11 25 20 45 0.556 12 23 8 31 0.742 13 19 16 35 0.543 14 18 19 37 0.486 15 17 20 37 0.459 16 18 27 45 0.400 17 19 14 33 0.576 18 16 17 33 0.485 19 18 19 37 0.486 20 15 23 38 0.395 21 17 15 32 0.531 22 19 19 38 0.500 23 18 27 45 0.400 24 17 11 28 0.607 25 14 17 31 0.452 26 22 17 39 0.564 27 18 25 43 0.419Explanation / Answer
Confidence interval mean proportion, p 0.4993 sample size, n 27 For 99% CI, t-value is 2.7787 Standard Error, SE = sqrt(p*(1-p)/n) 0.0962 Margin of error, ME = t*SE 0.2674 Lower limit = p - ME 0.2319 Upper Limit = p + ME 0.7666 Confidence interval (0.2319,0.7666) class 2 mean proportion, p 0.5351 sample size, n 26 For 99% CI, t-value is 2.7874 Standard Error, SE = sqrt(p*(1-p)/n) 0.0978 Margin of error, ME = t*SE 0.2727 Lower limit = p - ME 0.2624 Upper Limit = p + ME 0.8077 Confidence interval (0.2624,0.8077)
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