A researcher examines 28 sedimentary samples for iron concentration. The mean ir

Question

A researcher examines 28 sedimentary samples for iron concentration. The mean iron concentration for the sample data is 0.729 cc/cubic meter with a standard deviation of 0.0439. Determine the 95 % confidence interval for the population mean iron concentration. Assume the population is approximately normal. population mean ron concentratin r Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places. Answer(How to Enter) 2 Points Keypad

Explanation / Answer

TRADITIONAL METHOD
given that,
sample mean, x =0.729
standard deviation, s =0.0439
sample size, n =28
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.0439/ sqrt ( 28) )
= 0.008
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 27 d.f is 2.052
margin of error = 2.052 * 0.008
= 0.017
III.
CI = x ± margin of error
confidence interval = [ 0.729 ± 0.017 ]
= [ 0.712 , 0.746 ]
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DIRECT METHOD
given that,
sample mean, x =0.729
standard deviation, s =0.0439
sample size, n =28
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 27 d.f is 2.052
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 = [ 0.729 ± t a/2 ( 0.0439/ Sqrt ( 28) ]
= [ 0.729-(2.052 * 0.008) , 0.729+(2.052 * 0.008) ]
= [ 0.712 , 0.746 ]
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interpretations:
1) we are 95% sure that the interval [ 0.712 , 0.746 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

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