Solution:

From the given information, \(\displaystyle{n}={211},\ \overline{{{x}}}={33.7}{h}{g}\) and \(\displaystyle{s}={6.8}{h}{g}.\)

Step 2

The confidence level is 0.90.

\(\displaystyle{1}-\alpha={0.90}\)

\(\displaystyle\alpha={1}-{0.90}\)

\(\displaystyle\alpha={0.10}\)

Then, a 90% confidence interval for the population mean is

\(\displaystyle\overline{{{x}}}\pm{t}_{{{\frac{{\alpha}}{{{2}}}},\ {n}-{1}}}{\left({\frac{{{s}}}{{\sqrt{{{n}}}}}}\right)}={33.7}\pm{t}_{{{\frac{{{0.10}}}{{{2}}}},\ {211}-{1}}}{\left({\frac{{{6.8}}}{{\sqrt{{{211}}}}}}\right)}\)

\(\displaystyle={33.7}\pm{t}_{{{0.05},\ {210}}}{\left({\frac{{{6.8}}}{{\sqrt{{{211}}}}}}\right)}\)

\(\displaystyle={33.7}\pm{\left({1.625}\right)}{\left({\frac{{{6.8}}}{{\sqrt{{{211}}}}}}{\left[\text{Using the excel function}\ ={T}.{I}{N}{V}{\left({0.05},\ {210}\right)}\right]}\right.}\)

\(\displaystyle={33.7}\pm{0.7734}\)

\(\displaystyle={\left({33.7}-{0.7734},\ {33.7}+{0.07734}\right)}\)

\(\displaystyle={\left({32.9},\ {34.5}\right)}\)

Thus, a 90% confidence interval for the population mean is \(\displaystyle{32.9}{h}{g}{ < }\mu{ < }{34.5}{h}{g}\)

Step 3

From the given information, the confidence interval for the population mean is \(\displaystyle{32.1}{h}{g}{ < }\mu{ < }{34.3}{h}{g}\)

From the above two confidence intervals it can be observed that both lower and upper limits of the two confidence intervals are different.

Step 4

Reason for the incorrect options:

From the two confidence intervals it can be observed that limits are not same. To determine the similarity of the two confidence intervals it has to the limits of the intervals.

Thus, the options B, C and D are incorrect.