In one study, scientists have discovered increased levels of the hormone adrenoc

Question

In one study, scientists have discovered increased levels of the hormone adrenocorticotropin in people just before they awake from sleeping. In the study, 15 subjects were monitored during their sleep after being told they would be woken at a particular time. One hour prior to the designated wake- up time, the adrenocorticotropin level (in pg/mL) was measured in each, with the following results: mu = 37.3, sigma = 13.9. What is the point estimate for calculating this confidence interval? What is the critical value (indicate whether it is a z- or t- critical value) for a 95% confidence interval? Find the margin of error for a 95% confidence interval Give the 95% confidence interval Give a statement of confidence

Explanation / Answer

a.
point of estimate = 37.3
b.
critical value = 1.96
c.
Margin of Error = Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
Mean(x)=37.3
Standard deviation( sd )=13.9
Sample Size(n)=15
Margin of Error = Z a/2 * 13.9/ Sqrt ( 15)
= 1.96 * (3.59)
= 7.03
d.
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=37.3
Standard deviation( sd )=13.9
Sample Size(n)=15
Confidence Interval = [ 37.3 ± Z a/2 ( 13.9/ Sqrt ( 15) ) ]
= [ 37.3 - 1.96 * (3.59) , 37.3 + 1.96 * (3.59) ]
= [ 30.27,44.33 ]
e.
Interpretations:
1) We are 95% sure that the interval [30.27 , 44.33 ] 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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