M10Q1 Part3: A. Express the confidence interval (22.3%,33.3%) in the form of p^±
ID: 3243124 • Letter: M
Question
M10Q1
Part3:
A. Express the confidence interval (22.3%,33.3%) in the form of p^±ME.
_________% ±__________%
B. Express the confidence interval 18.3% < p < 33.7% in the form of p^±ME.
_________% ±________%
C. Express the confidence interval 51.4%±8% in the form of a trilinear inequality.
__________%
D. You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proportion is approximately p=34%. You would like to be 99% confident that your esimate is within 1% of the true population proportion. How large of a sample size is required?______________
E. You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable estimate for the population proportion. You would like to be 95% confident that you esimate is within 4% of the true population proportion. How large of a sample size is required?
n =______________
F. Assume that a sample is used to estimate a population proportion p. Find the margin of error M.E. that corresponds to a sample of size 276 with 25% successes at a confidence level of 99.9%. M.E.=________%
Explanation / Answer
a) here a= 22.3 , b = 33.3 (i am neglecting % here)
p^ = (a+b)/2 = 27.8
ME = 27.8-22.3 = 5.5
b) a =18.3 , b =33.7
p^ = 26
ME = 7.7
c) 51.4%±8% in the form of a trilinear inequality.
= 51.4 - 8 < p < 51.4+8
= 43.4 < p<59.4
d) n = (z^2 * pq)/(e^2)
here z = 2.576 for 99 % confidence
p = 0.34 ,q = 0.66
e = 0.01
hence n = 2.576^2 * 0.34*0.66/0.01^2 =
14890.6813
F) if p^ is not known
n = z^2 )/(4*e^2)
for 95 % z= 1.96
e = 0.04
hence n = 1.96^2 /(4*0.04^2) = 600.25
n= 601
F) ME = z*sqrt(p^q^/n)
for 99.9 % z = 3.291
p^ = 0.25
n = 276
hence ME = 3.291 * sqrt(0.25*0.75/276) =0.0857776
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