Given

\(\displaystyle\overline{{x}}={45.2}\)

n=31

\(\displaystyle\sigma={5.3}\)

Determine the t-value by looking in the row starting with degrees of freedom df=n-1=31-1=30 and in the column with c=90%/95%/99% in table 4:

\(\displaystyle{90}\%:{t}_{{{\frac{{\alpha}}{{{2}}}}}}={1.697}\)

\(\displaystyle{95}\%:{t}_{{{\frac{{\alpha}}{{{2}}}}}}={2.042}\)

\(\displaystyle{99}\%:{t}_{{{\frac{{\alpha}}{{{2}}}}}}={2.750}\)

The confidence interval is:

\(\displaystyle\overline{{x}}-{t}_{{{\frac{{\alpha}}{{{2}}}}}}\dot{{\lbrace}}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\ \to\ \overline{{x}}+{t}_{{{\frac{{\alpha}}{{{2}}}}}}\dot{{{\frac{{{s}}}{{\sqrt{{{n}}}}}}}}\)

Fill in the known values: NSL \(\displaystyle{45.2}-{t}_{{{\frac{{\alpha}}{{{2}}}}}}\dot{{\lbrace}}{\frac{{{5.3}}}{{\sqrt{{{31}}}}}}\ \to\ {45.2}+{t}_{{{\frac{{\alpha}}{{{2}}}}}}\dot{{\lbrace}}{\frac{{{5.3}}}{{\sqrt{{{31}}}}}}\)

Simplify:

90%:43.5846 to 46.8154

95%:43.2562 to 47.1438

99%:42.5823 to 47.8177

\(\displaystyle\overline{{x}}={45.2}\)

n=31

\(\displaystyle\sigma={5.3}\)

Determine the t-value by looking in the row starting with degrees of freedom df=n-1=31-1=30 and in the column with c=90%/95%/99% in table 4:

\(\displaystyle{90}\%:{t}_{{{\frac{{\alpha}}{{{2}}}}}}={1.697}\)

\(\displaystyle{95}\%:{t}_{{{\frac{{\alpha}}{{{2}}}}}}={2.042}\)

\(\displaystyle{99}\%:{t}_{{{\frac{{\alpha}}{{{2}}}}}}={2.750}\)

The confidence interval is:

\(\displaystyle\overline{{x}}-{t}_{{{\frac{{\alpha}}{{{2}}}}}}\dot{{\lbrace}}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\ \to\ \overline{{x}}+{t}_{{{\frac{{\alpha}}{{{2}}}}}}\dot{{{\frac{{{s}}}{{\sqrt{{{n}}}}}}}}\)

Fill in the known values: NSL \(\displaystyle{45.2}-{t}_{{{\frac{{\alpha}}{{{2}}}}}}\dot{{\lbrace}}{\frac{{{5.3}}}{{\sqrt{{{31}}}}}}\ \to\ {45.2}+{t}_{{{\frac{{\alpha}}{{{2}}}}}}\dot{{\lbrace}}{\frac{{{5.3}}}{{\sqrt{{{31}}}}}}\)

Simplify:

90%:43.5846 to 46.8154

95%:43.2562 to 47.1438

99%:42.5823 to 47.8177