Mathematics, 23.07.2020 19:01 sdfhghjg

The variance in drug weights is critical in the pharmaceutical industry. For a specific drug, with weights measured in grams, a sample of 18 units provided a sample variance of 0.36. a. Construct a 90% confidence interval estimate of the population variance for the weight of this drug. Show your work. b. Construct a 90% confidence interval estimate of the population standard deviation.

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Answer from: hiji0206

a) 0.2218 to 0.7057

b) 0.4710 to 0.8401

Step-by-step explanation:

Given:

sample = n = 18

sample variance = s² = 0.36

To find:

a) 90% confidence interval estimate of the population variance.

b) 90% confidence interval estimate of the population standard deviation.

Compute degree of freedom

degree of freedom = df = n - 1 = 18 - 1 = 17

Compute value of α for a 90% confidence interval

The confidence level for 90% is:

c = 0.90

α = 1 - c = 1 - 0.90 = 0.1

α = 0.1

Now use the table of critical values of the chi-square distribution in order to find critical values. Search for the 17th row of table using df = 17 and find the column corresponding to 1 - α/2 i.e. 0.95 of table for upper tail critical values and column corresponding to α /2 i.e. 0.05 of table for lower tail critical values.

Using the table:

$X^{2}_{0.95}$ = 8.672

$X^{2}_{0.05}$ = 27.587

90% Confidence interval estimate of the population variance

The boundaries of CI are computed using formula:

(n−1) s² / $X^{2}_{\alpha/2}$ ≤ σ² ≤ (n−1) s² / $X^{2}_{1-\alpha/2}$

(n−1) s² / $X^{2}_{\alpha/2}$ = (18-1) 0.36 / 27.587

= (17) 0.36 / 27.587

= 6.12 / 27.587

(n−1) s² / $X^{2}_{\alpha/2}$ = 0.2218

(n−1) s² / $X^{2}_{1-\alpha/2}$ = (18-1) 0.36 / 8.672

= (17) 0.36 / 8.672

= 6.12 / 8.672

= 0.7057

This results in inequalities 0.2218 ≤ σ² ≤ 0.7057 for the variance

90% Confidence interval estimate of the population standard deviation

The boundaries of CI are computed using formula:

√(n−1) s² / $X^{2}_{\alpha/2}$ ≤ σ ≤ √(n−1) s² / $X^{2}_{1-\alpha/2}$

√(n−1) s² / $X^{2}_{\alpha/2}$ = √((18-1) 0.36 / 27.587)

= √((17) 0.36 / 27.587)

= √(6.12 / 27.587)

= √0.2218

= 0.4709

= 0.4710

√(n−1) s² / $X^{2}_{1-\alpha/2}$ = √((18-1) 0.36 / 8.672)

= √((17) 0.36 / 8.672)

= √ (6.12 / 8.672)

= √0.7057

= 0.8401

0.4710 ≤ σ ≤ 0.8401 for the standard deviation.

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The variance in drug weights is critical in the pharmaceutical industry. For a specific drug, with w...