Answer:
[tex]220-2.01\frac{12}{\sqrt{50}}=216.589[/tex]
[tex]220+2.01\frac{12}{\sqrt{50}}=223.411[/tex]
The confidence interval for this case would be (216.589, 223.411)
Step-by-step explanation:
Information given
[tex]\bar X=220[/tex] represent the sample mean
[tex]\mu[/tex] population mean
s=12 represent the sample standard deviation
n=50 represent the sample size
Confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom are given by:
[tex]df=n-1=50-1=49[/tex]
The Confidence level is 0.95 or 95%, the significance is [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], the critical value for this case is [tex]t_{\alpha/2}=2.01[/tex]
And replacing we got:
[tex]220-2.01\frac{12}{\sqrt{50}}=216.589[/tex]
[tex]220+2.01\frac{12}{\sqrt{50}}=223.411[/tex]
The confidence interval for this case would be (216.589, 223.411)
