Carbohydrate measured on one more sample was $\mathbf{12}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{wt}\mathbf{\%}$Using six results, find the $\mathbf{90}\mathbf{\%}$confidence interval

Mean $\overline{x}=\frac{{\displaystyle \underset{\mathrm{i}}{\sum {\mathrm{x}}_{\mathrm{i}}}}}{\mathrm{n}}.$

SD $\mathrm{s}=\sqrt{\frac{{\displaystyle \underset{\mathrm{i}}{\sum {\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^2}}}{\mathrm{n}-1}}.$

Confidenceinterval$=\overline{\mathrm{x}}\pm \frac{\mathrm{ts}}{\sqrt{\mathrm{n}}}.$

The carbohydrate content of glycoprotein is listed below:

$12.6,11.9,13.0,12.7,12.5,{}_{\mathrm{and}}12.3\mathrm{wt}\%.$

Calculation of the confidence interval:

The number of measurements $\left(\mathrm{n}\right)-6.$

The degrees of freedom $\left(n-1\right)=5.$

Calculate the mean and the standard deviation for given measurements of the carbohydrate content of a glycoprotein.

Mean:

$\overline{x}=\frac{{\displaystyle \underset{\mathrm{i}}{\sum {\mathrm{x}}_{\mathrm{i}}}}}{\mathrm{n}}$

$=\frac{12.6+11.9+13.0+12.7+12.5+12.3}{6}$

$=\frac{75}{6}$

$=12.5.$

Standard deviation:

localid="1663331170102" $$\begin{gathered} \mathrm{s}=\sqrt{\frac{{\displaystyle \underset{\mathrm{i}}{\sum {\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^2}}}{\mathrm{n}-1}} \\ =\sqrt{\frac{{\left(12.5-12.6\right)}^2+.......+{\left(12.3-12.6\right)}^2}{6-1}} \\ =0.37. \end{gathered}$$

Confidence interval:

For $90\%$of the confidence interval, look for the localid="1663331293495" $\mathit{t}$ value in Table $4-4,$, across five degrees of freedom.

The $\mathit{t}$value is $2.015.$

Therefore , the $90\%$confident interval

$$\begin{gathered} =\overline{\mathrm{x}}\pm \frac{\mathrm{ts}}{\sqrt{\mathrm{n}}} \\ =12.5\pm \frac{\left(2.015\right)\left(0.37\right)}{\sqrt{6}} \\ =12.5\pm \frac{0.74555}{2.449} \\ =12.5\pm 0.3\mathrm{wt}\% \end{gathered}$$

Hence, 90% confident interval is$12.5\pm 0.3\mathrm{wt}\%.$