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# 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.

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Confidence intervals
asked 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.

## Answers (1)

2020-12-04
Step 1 Data values are given population standard deviation is not known hence we use T Interval. Sample size $$\displaystyle{\left({n}\right)}={5}$$ Degrees of freedom(df): $$\displaystyle{d}{f}={n}-{1}$$
$$\displaystyle={5}-{1}$$
$$\displaystyle={4}$$ Confidence level =0.90 Use T table or excel command =T.INV.2T $$\displaystyle{\left({1}-{0.90},{4}\right)}$$ to get critical value Critical value $$\displaystyle{\left({t}_{{{c}}}\right)}={2.132}$$ $$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}&{X}&{\left({X}−{12.54}\right)}^{{{2}}}\backslash{h}{l}\in{e}&{12.6}&{0.0036}\backslash{h}{l}\in{e}&{11.9}&{0.4096}\backslash{h}{l}\in{e}&{13}&{0.2116}\backslash{h}{l}\in{e}&{12.7}&{0.0256}\backslash{h}{l}\in{e}&{12.5}&{0.0016}\backslash{h}{l}\in{e}{T}{o}{t}{a}{l}&{62.7}&{0.652}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ Sample mean: $$\displaystyle\overline{{{x}}}={\frac{{{\sum_{{{i}={1}}}^{{5}}}\xi}}{{{n}}}}$$
$$\displaystyle={\frac{{{62.7}}}{{{5}}}}$$
$$\displaystyle{12.54}$$ Sample standard deviation: $$\displaystyle{s}=\sqrt{{{\frac{{{\sum_{{{i}={1}}}^{{5}}}{\left({x}_{{{i}}}-\overline{{{x}}}\right)}^{{{2}}}}}{{{n}-{1}}}}}}$$
$$\displaystyle=\sqrt{{{\frac{{{\sum_{{{i}={1}}}^{{5}}}{\left({x}_{{{i}}}-{12.54}\right)}^{{{2}}}}}{{{5}-{1}}}}}}$$
$$\displaystyle=\sqrt{{{\frac{{{0.652}}}{{{4}}}}}}$$
$$\displaystyle=\sqrt{{{0.163}}}$$
$$\displaystyle={0.403733}$$ Step 3 $$\displaystyle=\overline{{{x}}}\pm{t}_{{{c}}}{\frac{{{s}}}{{\sqrt{{{n}}}}}}$$
$$\displaystyle={12.54}\pm{2.132}\times{\frac{{{0.403733}}}{{\sqrt{{{5}}}}}}$$
$$\displaystyle={12.54}\pm{2.132}\times{0.180555}$$
$$\displaystyle={12.54}\pm{0.3849}$$
$$\displaystyle={\left({12.54}-{0.3849},{12.54}+{0.3849}\right)}$$
$$\displaystyle={\left({12.1551},{12.9249}\right)}$$ Step 4 Answer: 90% confidence interval is ( 12.16, 12.92 )

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