Given information-

Sample mean, \(\displaystyle{x}-\overline{=}{7.1}\)

Population variance, \(\displaystyle\sigma{2}={0.6084}\)

So, Standard deviation \(\displaystyle={0.78}\)

Confidence level, \(\displaystyle{c}={99}\%\)

Sample size, \(\displaystyle{n}={10}\)

Significance level \(\displaystyle{\left(\alpha\right)}={1}-{0.99}={0.01}\)

Step 2

Since here, population variance is given, therefore using z-test.

99% Confidence interval is given by the following formula

\(\displaystyle{C}.{I}=\overline{{{x}}}\pm{z}_{{{\frac{{{0.1}}}{{{2}}}}}}\times{\frac{{\sigma}}{{\sqrt{{{n}}}}}}\)

\(\displaystyle{C}.{I}={7.1}\pm{2.57583}\times{\frac{{{0.78}}}{{\sqrt{{{10}}}}}}={\left({6.4646},{7.7353}\right)}\)

\(\displaystyle{z}_{{{\frac{{{0.01}}}{{{2}}}}}}=\pm{2.57583}\) (Using excel =NORM.S.INV (0.005))

Since sample mean is in the range of confidence interval, therefore the sample mean is significant at the given significance level.