st norbert\'s college in green bay, wisconsin and wisconsin public radio conduct

Question

st norbert's college in green bay, wisconsin and wisconsin public radio conduct an annual poll of wisconsinites about political opinions. The dall 2011 survey asked a random sample of 402 adult wisconsin residents whether they think things in the country are going in the right direction or in the wrong direction. 66% said that things were going in the wrong direction.

A) calculate the margin of error for the proportion of a;; adult wisconsin residents who think things are going in the wrong direction for 90% confidence.

B) Would the margin of error be larger or smaller for 95% confidence? Explain.

Are the assumptions and conditions met?

How many people would need to be surveyed for a 90% confident interval to ensure the margin of error would be less than 2%

Explanation / Answer

a)
Margin of Error = Z a/2 Sqrt(p*(1-p)/n))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Sample Size(n)=402
Sample proportion =0.66
Margin of Error = Z a/2 * ( Sqrt ( (0.66*0.34) /402) )
= 1.64* Sqrt(0.001)
=0.039
b)
WITH 95% C.I
Margin of Error = Z a/2 * ( Sqrt ( (0.66*0.34) /402) )
= 1.96* Sqrt(0.001)
=0.046
It is larger
c)
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.1 is = 1.64
Samle Proportion = 0.66
ME = 0.02
n = ( 1.64 / 0.02 )^2 * 0.66*0.34
= 1508.866 ~ 1509

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