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  3. 2. it is estimated that the standard deviation of

2. It is estimated that the standard deviation of the height of American adult males is...

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Basic Math

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 Solved by verified expert
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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