2. It is estimated that the standard deviation of the height
of American adult males is...
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2. It is estimated that the standard deviation of the heightof American adult males is 4 inches. Suppose we survey a sample of50 men and find a sample mean of 69 inches.
(a) For a confidence level of %90, what is the margin of errorand the confidence interval?
(b) For a confidence level of %95, what is the margin of errorand the confidence interval?
(c) If we sample 500 men instead, and still insist on %90confidence level, what is the new margin of error and the newconfidence interval?
(d) If we wanted a 95% confidence level and we wanted a marginof error less than 0.1, how many men would we need to survey (whatmust our sample size be)?
Answer & Explanation
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4.3 Ratings (874 Votes)
a)
sample mean
x?=
69.0000
sample size
n=
50.00
population std deviation
?=
4.000
standard errror of mean =
?x=?/?n=
0.5657
for 90 % CI value of z=
1.645
margin of
error E=z*std
error
=
0.930
lower
confidence bound=sample mean-margin of error=
68.070
Upper
confidence bound=sample mean +margin of error=
69.930
b)
for 95 % CI value of z=
1.960
margin of
error E=z*std
error
=
1.109
lower
confidence bound=sample mean-margin of error=
67.891
Upper
confidence bound=sample mean +margin of error=
70.109
c)
standard errror of mean =
?x=?/?n=
0.1789
for 90 % CI value of z=
1.645
margin of
error E=z*std
error
=
0.294
lower
confidence bound=sample mean-margin of error=
68.706
Upper
confidence bound=sample mean +margin of error=
69.294
d)
for95% CI crtiical
Z =
1.96
standard deviation ?=
4.000
margin of error E =
0.1
required
sample size
n=(z?/E)2
=
6147
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