Answer:
a)
![6.8 -2.110*1.2 =4.268](https://img.qammunity.org/2021/formulas/mathematics/college/jof5ni6xh18k4ayflyt9x11slyef3qj34l.png)
![6.8 +2.110*1.2 =9.332](https://img.qammunity.org/2021/formulas/mathematics/college/rlw3bj25ajwkie80b6qqjn1prpekguy88t.png)
b)
![ME = t_(\alpha/2) SE= 2.110*1.2= 2.532](https://img.qammunity.org/2021/formulas/mathematics/college/2gjyw42nq3ei2m7apdxf1gwu1e68e751bq.png)
Explanation:
For this case we have the following info given:
represent the sample mean
represent the standard error
the sample size
Part a
For this case the confidence interval for the mean is given by:
![\bar X \pm t_(\alpha/2) SE](https://img.qammunity.org/2021/formulas/mathematics/college/mts415zdoj31tumatl6e00y5ejimb0z17o.png)
The degrees of freedom are given by:
![df =n-1=18-1=17](https://img.qammunity.org/2021/formulas/mathematics/college/rmc8jh13bpj87zfbl5zuoc7wtapj4wb7ww.png)
And the critical value for a confidence interval of 95% is given by:
![t_(\alpha/2)=2.110](https://img.qammunity.org/2021/formulas/mathematics/college/sd0hri2n19x03dxptqoo1bxhlht97a1902.png)
And the confidence interval would be:
![6.8 -2.110*1.2 =4.268](https://img.qammunity.org/2021/formulas/mathematics/college/jof5ni6xh18k4ayflyt9x11slyef3qj34l.png)
![6.8 +2.110*1.2 =9.332](https://img.qammunity.org/2021/formulas/mathematics/college/rlw3bj25ajwkie80b6qqjn1prpekguy88t.png)
Part b
The margin of error is given by:
![ME = t_(\alpha/2) SE= 2.110*1.2= 2.532](https://img.qammunity.org/2021/formulas/mathematics/college/2gjyw42nq3ei2m7apdxf1gwu1e68e751bq.png)