Answer:
The degrees of freedom are given by:
[tex]df=n-1=317-1=316[/tex]
And replaicing we got:
[tex]29-2.1=26.9[/tex] Â Â
[tex]29+2.1=31.1[/tex] Â Â
The 95% confidence interval would be between 26.9 and 31.1
Step-by-step explanation:
Information given
[tex]\bar X= 29[/tex] represent the sample mean
[tex]\mu[/tex] population mean
s represent the sample standard deviation
[tex] ME= 2.1[/tex] represent the margin of error
n represent the sample size Â
Solution
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] Â (1)
And this formula is equivalent to:
[tex] \bar X \pm ME[/te]x
The degrees of freedom are given by:
[tex]df=n-1=317-1=316[/tex]
And replaicing we got:
[tex]29-2.1=26.9[/tex] Â Â
[tex]29+2.1=31.1[/tex] Â Â
The 95% confidence interval would be between 26.9 and 31.1