Answers to Practice Problems in Chap. 4, Use of Statistics in Food Analysis

Abstract

What is the mean of the observations (mg/cup)?x¯=∑xin=x1+x2+…+xn−1+xnn=1967.1mg/cup6=327.85mg/cup,roundto328mg/cup$$ \overline{x} = \frac{{\displaystyle \sum }{x}_i}{n}=\frac{x_1+{x}_2+\dots +{x}_{n-1}+{x}_n}{n}=\frac{1967.1\ \mathrm{mg}/\mathrm{cup}}{6}=327.85\ \mathrm{mg}/\mathrm{cup},\kern0.5em \mathrm{round}\ \mathrm{to}\ 328\ \mathrm{mg}/\mathrm{cup} $$What is the standard deviation of the observations (mg/cup)? Since we have < 30 observations, the correct formula for sample standard deviation is:SDn−1=∑(xi−x¯)2n−1=∑(xi−327.85mg/cup)26−1=460.215mg2/cup25=9.593904mg/cup$$ {\mathrm{SD}}_{n-1}=\sqrt{\frac{{\displaystyle \sum }{\left({x}_i-\overline{x}\ \right)}^2}{n-1}}=\sqrt{\frac{{\displaystyle \sum }{\left({x}_i-327.85\ \mathrm{mg}/\mathrm{cup}\right)}^2}{6-1}} = \sqrt{\frac{460.215\ {\mathrm{mg}}^2/{\mathrm{cup}}^2}{5}} = 9.593904\ \mathrm{mg}/\mathrm{cup} $$Calculate the 96 % confidence interval for the true population mean (use t-score, not Z score, because n is small). We know that the formula for a CI is:CI:x¯±tα2,df=n−1×SDn$$ \mathrm{CI}:\kern1.25em \overline{x} \pm {t}_{\frac{\upalpha}{2},\kern0.5em \mathrm{df}= n-1} \times \frac{\mathrm{SD}}{\sqrt{n}} $$We know the mean, SD, and n already, so we just need the t-score. First, calculate df:df=n−1=6−1=5$$ \mathrm{df}=