Answer:
a) [tex] 6.8 -2.110*1.2 =4.268[/tex]
[tex] 6.8 +2.110*1.2 =9.332[/tex]
b) [tex] ME = t_{\alpha/2} SE= 2.110*1.2= 2.532[/tex]
Step-by-step explanation:
For this case we have the following info given:
[tex] \bar X= 6.8[/tex] represent the sample mean
[tex] Se= 1.2[/tex] represent the standard error
[tex] n =18[/tex] the sample size
Part a
For this case the confidence interval for the mean is given by:
[tex] \bar X \pm t_{\alpha/2} SE[/tex]
The degrees of freedom are given by:
[tex] df =n-1=18-1=17[/tex]
And the critical value for a confidence interval of 95% is given by:
[tex] t_{\alpha/2}=2.110[/tex]
And the confidence interval would be:
[tex] 6.8 -2.110*1.2 =4.268[/tex]
[tex] 6.8 +2.110*1.2 =9.332[/tex]
Part b
The margin of error is given by:
[tex] ME = t_{\alpha/2} SE= 2.110*1.2= 2.532[/tex]