Step 1

Given:

Sample Size \(\displaystyle{\left({n}\right)}={49}\)

Sample Mean \(\displaystyle{\left(\overline{{X}}\right)}={72}\)

Population standard deviation \(\displaystyle{\left(\sigma\right)}={13.1}\)

Step 2

90% Confidence interval \(\displaystyle=\overline{{X}}\pm{Z}_{{\frac{\alpha}{{2}}}}{\left(\frac{\sigma}{\sqrt{{n}}}\right)}\)

\(\displaystyle={72}\pm{1.165}{\left(\frac{13.1}{\sqrt{{49}}}\right)}\)

\(\displaystyle={72}\pm{3.0785}\)

\(\displaystyle={\left({68.9215},{75.0785}\right)}\)

\(\displaystyle={\left({68.92},{75.08}\right)}\)

Since area corresponding to z value is given in z-table as:

\(\displaystyle{Z}_{{\frac{\alpha}{{2}}}}={Z}_{{\frac{0.10}{{2}}}}\)

\(\displaystyle={Z}_{{{0.05}}}\)

\(= 1.645\)

Given:

Sample Size \(\displaystyle{\left({n}\right)}={49}\)

Sample Mean \(\displaystyle{\left(\overline{{X}}\right)}={72}\)

Population standard deviation \(\displaystyle{\left(\sigma\right)}={13.1}\)

Step 2

90% Confidence interval \(\displaystyle=\overline{{X}}\pm{Z}_{{\frac{\alpha}{{2}}}}{\left(\frac{\sigma}{\sqrt{{n}}}\right)}\)

\(\displaystyle={72}\pm{1.165}{\left(\frac{13.1}{\sqrt{{49}}}\right)}\)

\(\displaystyle={72}\pm{3.0785}\)

\(\displaystyle={\left({68.9215},{75.0785}\right)}\)

\(\displaystyle={\left({68.92},{75.08}\right)}\)

Since area corresponding to z value is given in z-table as:

\(\displaystyle{Z}_{{\frac{\alpha}{{2}}}}={Z}_{{\frac{0.10}{{2}}}}\)

\(\displaystyle={Z}_{{{0.05}}}\)

\(= 1.645\)