Question

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:____________ .

Answer 1

`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)