Question

A 1-ounce serving of a certain breakfast cereal is supposed to contain 9 grams of psyllium, a high fiber food product that may be beneficial in lowering cholesterol levels. Twenty-four servings were analyzed for the amount of psyllium, and the sample standard deviation was 0.52. Construct a 90% confidence interval for the true standard deviation of the amount of psyllium per serving of this cereal. Write a sentence summarizing your results.

Answer 1

Solution :

Given that,

s = 0.52

s² = 0.2704

n = 9

Degrees of freedom = df = n - 1 = 9 - 1 = 8

At 90% confidence level the ² value is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

1 - / 2 = 1 - 0.05 = 0.95

²_L = ²_/2,df = 15.507

²_R = ²_{1 - /2,df} = 2.733

The 90% confidence interval for is,

(n - 1)s² / ²_/2 < < (n - 1)s² / ²_{1 - /2}

8 * 0.2704 / 15.507 < < 8 * 0.2704 / 15.507

0.37 < < 0.89

(0.37 , 0.89)