ankarskogC

2020-12-03

In replicate analyses, the carbohydrate content of a glycoprotein ( a protein with sugars attached to it) is found to be 12.6, 11.9, 13.0, 12.7, and 12.5 g of carbohydrate per 100 g of protein. Find the 90% confidence intervals for the carbohydrate content.

grbavit

Step 1 Data values are given population standard deviation is not known hence we use T Interval. Sample size $\left(n\right)=5$ Degrees of freedom(df): $df=n-1$
$=5-1$
$=4$ Confidence level $=0.90$ Use T table or excel command =T.INV.2T $\left(1-0.90,4\right)$ to get critical value Critical value $\left({t}_{c}\right)=2.132$

$\begin{array}{|ccc|}\hline & X& \left(X-12.54{\right)}^{2}\\ & 12.6& 0.0036\\ & 11.9& 0.4096\\ & 13& 0.2116\\ & 12.7& 0.0256\\ & 12.5& 0.0016\\ Total& 62.7& 0.652\\ \hline\end{array}$

Sample mean: $\stackrel{―}{x}=\frac{\sum _{i=1}^{5}\xi }{n}$
$=\frac{62.7}{5}$
$12.54$ Sample standard deviation: $s=\sqrt{\frac{\sum _{i=1}^{5}{\left({x}_{i}-\stackrel{―}{x}\right)}^{2}}{n-1}}$
$=\sqrt{\frac{\sum _{i=1}^{5}{\left({x}_{i}-12.54\right)}^{2}}{5-1}}$
$=\sqrt{\frac{0.652}{4}}$
$=\sqrt{0.163}$
$=0.403733$ Step 3 $=\stackrel{―}{x}±{t}_{c}\frac{s}{\sqrt{n}}$
$=12.54±2.132×\frac{0.403733}{\sqrt{5}}$
$=12.54±2.132×0.180555$
$=12.54±0.3849$
$=\left(12.54-0.3849,12.54+0.3849\right)$
$=\left(12.1551,12.9249\right)$

Step 4 Answer: $90\mathrm{%}$ confidence interval is ( 12.16, 12.92 )

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