Chapter 8: Q. 64 (page 518)

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:
$26 31 23 22 11 22 14 31$
Construct and interpret a $95 %$ confidence interval for the mean amount of vitamin C M in the CSB from this production run.

Short Answer

Expert verified

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

Step by step solution

Given information

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

$26 31 23 22 11 22 14 31$ .

Explanation

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

Determine the standard deviation as:

$s_{x} = \sqrt{\frac{\sum {(x_{i} - \bar{x})}^{2}}{n - 1}} = \sqrt{\frac{(26 - 22.5)^{2} + \dots + (31 - 22.5)^{2}}{8 - 1}} = 7.191$

Explanation

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

$d f = = 8 - 1 = 7$

Convert , the confidence level is $95 %$ into decimal:
$\frac{95}{100} = 0.95$
Determine the column as:
$\frac{1 - c}{2} = \frac{1 - 0.95}{2} = 0.025$
Calculate the critical value $t^{*}$ , from the table $B$ use the row $d f$ 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:

\bar{x} - E < μ < \bar{x} + E

$22.5 - 6.0128 < μ < 22.5 + 6.0128$

$16.4872 < μ < 28.5128$

Hence,

95 %

confident of the population mean is between

16.4872 mg / 100 g

and

16.4872 mg / 100 g

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