Question

1.     WHAT IS THE MARGIN OF ERROR FOR TH POPULATION MEAN AT 90% CONFIDENCE LEVEL FOR THE INFORMATION   OK DONE

_____ 0.75

______ 0.74

_______ 0.72

______ NONE OF THE ANSWER MATCH MY CALCULATIONS

_____ 0.73

Answer 1

1.

Assumed data,
not given in the data,
given that,
sample mean, x =67
standard deviation, s =2.5
sample size, n =34

TRADITIONAL METHOD
given that,
sample mean, x =67
standard deviation, s =2.5
sample size, n =34
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 2.5/ sqrt ( 34) )
= 0.43
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 33 d.f is 1.69
margin of error = 1.69 * 0.43
= 0.72
III.
CI = x ± margin of error
confidence interval = [ 67 ± 0.72 ]
= [ 66.28 , 67.72 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =67
standard deviation, s =2.5
sample size, n =34
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 33 d.f is 1.69
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 67 ± t a/2 ( 2.5/ Sqrt ( 34) ]
= [ 67-(1.69 * 0.43) , 67+(1.69 * 0.43) ]
= [ 66.28 , 67.72 ]
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interpretations:
1) we are 90% sure that the interval [ 66.28 , 67.72 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean

Answer:
margin of error =0.72
option:C