Answer:
Following are the solution to this question
Step-by-step explanation:
Please find the complete question in the attached file.
In point a:
[tex]a=0.05 \\\\| Z(0.025)|= 1.96 \text{(check in normal table)} \\\\ 95 \% \ CI \\\\\bar x \pm \ Z \times \frac{s}{\sqrt{n}}\\\\\to 80 \pm 1.96 \times \frac{15}{\sqrt{60}} \\\\\to (76.20448, 83.79552)[/tex]
In point b:
[tex]95 \% \ CI \ is \\\\ \bar x \pm Z \times \frac{s}{\sqrt{n}}\\\\\to 80 \pm 1.96 \times \frac {15}{\sqrt{120}}\\\\\to (77.31616, 82.68384)[/tex]
In point c:
A large sample is a smaller error margin.