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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 100grams (mg/100 g) of blend, for a random sample of size8 from one production run:
2631232211221431
Construct and interpret a95%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 16.4872mg/100mgand 28.5128mg/100mg, according to 95%confidence.

Step by step solution

01

Given information

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

2631232211221431.

02

Explanation

Determine the mean as:
x炉=sumofthesamplestotal number ofsample=26+31+23+22+11+22+14+318=22.5

Determine the standard deviation as:

sx=鈭�xi-x炉2n-1=(26鈭�22.5)2+鈰�+(31鈭�22.5)28鈭�1=7.191

03

Explanation

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

df==8-1=7

Convert , the confidence level is95% into decimal:
95100=0.95
Determine the column as:
1-c2=1-0.952=0.025
Calculate the critical value t*, from the table Buse the rowdfand column. Hence, the critical value is t*=2.365.

04

Explanation

Determine the margin of error as:
E=t*sxn

=2.365脳7.1918

=6.0128

Finally, determine the population mean as:
x炉-E<渭<x炉+E

22.5-6.0128<渭<22.5+6.0128

16.4872<渭<28.5128

Hence, 95%confident of the population mean is between 16.4872mg/100gand 16.4872mg/100g.

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