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A sample of size =n48 has sample mean x=54.6 and sample standard deviation =s9.2. Part: 0 / 20 of 2 Parts Complete Part 1 of 2 Construct a 99.9% confidence interval for the population mean μ. Round the answers to one decimal place. A 99.9% confidence interval for the population mean is:____________ .

User Lazette
by
6.6k points

1 Answer

4 votes

Answer:


54.6-3.51(9.2)/(√(48))=49.94


54.6+3.51(9.2)/(√(48))=59.26

The confidence interval is given by (49.94, 59.26)

Explanation:

Info given


\bar X=54.6 represent the sample mean


\mu population mean (variable of interest)

s=9.2 represent the sample standard deviation

n=48 represent the sample size

Part a

The confidence interval for the mean is given by the following formula:


\bar X \pm t_(\alpha/2)(s)/(√(n)) (1)

The degrees of freedom are given by:


df=n-1=48-1=47

The Confidence is 0.999 or 99.9%, and the significance is
\alpha=0.001 and
\alpha/2 =0.0005, and the critical value would be
t_(\alpha/2)=3.51

And replacing we got:


54.6-3.51(9.2)/(√(48))=49.94


54.6+3.51(9.2)/(√(48))=59.26

The confidence interval is given by (49.94, 59.26)

User Adam Pearlman
by
7.3k points
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