Question

64. Vitamin C content Several years ago, the U.S. Agency for International Development provided $238,300$ metric tons of corn-soy blend (CSB) for
emergency relief in countries throughout the world. CSB is a highly nutritious, low-cost fortified food. As part of a study to evaluate appropriate vitamin C levels in this food, measurements were taken on samples of CSB produced in a factory. The following data are the amounts of vitamin C,

measured in milligrams per $100$grams (mg/100 g) of blend, for a random sample of size$8$ from one production run:
$2631232211221431$
Construct and interpret a$95\%$confidence interval for the mean amount of vitamin C M in the CSB from this production run.

Short answer

The population mean is between $16.4872mg/100mg$and $28.5128mg/100mg$, according to $95\%$confidence.

Step-by-step solution

Given information

The amounts of vitamin C, measured in milligrams per $100$grams $(mg/100g)$of blend, for a random sample of size $8$from one production run:

$2631232211221431$.

Explanation

Determine the mean as:
$$\begin{gathered} \overline{x}=\frac{\text{sumofthesamples}}{\text{total number ofsample}} \\ =\frac{26+31+23+22+11+22+14+31}{8} \\ =22.5 \end{gathered}$$

Determine the standard deviation as:

$$\begin{gathered} s_x=\sqrt{\frac{\sum {\left(x_i-\overline{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{(26-22.5{)}^2+\cdots +(31-22.5{)}^2}{8-1}} \\ =7.191 \end{gathered}$$

Explanation

Next, determine the degree of freedom as:
$df=n-1$

$df==8-1=7$

Convert , the confidence level is$95\%$ into decimal:
$\frac{95}{100}=0.95$
Determine the column as:
$$\begin{gathered} \frac{1-c}{2}=\frac{1-0.95}{2} \\ =0.025 \end{gathered}$$
Calculate the critical value $t^{*}$, from the table $B$use the row$df$and column. Hence, the critical value is $t^{*}=2.365$.

Explanation

Determine the margin of error as:
$E=t^{*}\frac{s_x}{\sqrt{n}}$

$=2.365\times \frac{7.191}{\sqrt{8}}$

$=6.0128$

Finally, determine the population mean as:
$\overline{x}-E<\mu <\overline{x}+E$

$22.5-6.0128<\mu <22.5+6.0128$

$16.4872<\mu <28.5128$

Hence, $95\%$confident of the population mean is between $16.4872\mathrm{mg}/100\mathrm{g}$and $16.4872\mathrm{mg}/100\mathrm{g}$.