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(15 points) A random sample of 80 Mizzou students showed that 44 had driven a car...

(15 points) A random sample of 80 Mizzou students showed that 44 had driven a car during the day before the survey was conducted. Suppose that we are interested in forming a 99.7 percent confidence interval for the proportion of all Mizzou students who drove a car the day before the survey was conducted.

a) The estimate is: .55

(b) The standard error is: .0556

(c) The multiplier is ????

C is not 1.645 according to text so what is it?

C is also not .165 according to text?

math   Statistics-And-Probability   
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Answer #1

as Estimate is 9 = 44 - 0.55 S.E(P) = /24 - 10.55x0.45 20.0556 80 Multiples : Confidence level = 0.997 2=1-0.987 = 0.003 .. 2

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