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

Answer:

$54.6-3.51\frac{9.2}{\sqrt{48}}=49.94$    

$54.6+3.51\frac{9.2}{\sqrt{48}}=59.26$    

The confidence interval is given by (49.94, 59.26)

Step-by-step 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:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (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:

$54.6-3.51\frac{9.2}{\sqrt{48}}=49.94$    

$54.6+3.51\frac{9.2}{\sqrt{48}}=59.26$    

The confidence interval is given by (49.94, 59.26)