Question

for a certain drug, based on standards set by the united states pharmacopeia (usp) - an official public standards - setting authority for all prescription and over - the - counter medicines and other health care products manufactured or sold in the united states, a standard deviation of capsule weights of less than 2 mg is acceptable. a sample of 12 capsules was taken and the weights are provided below:

the mean of the data set is 99.6; the sample standard deviation is 0.89; if the normality plot is not provided you may assume that the capsule weights are normally distributed.
construct a 90% confidence interval for the variance of all capsule weight.

Explanation:

Step1: Determine the formula for confidence interval of variance

The formula for the confidence interval of the population variance \(\sigma^{2}\) when the population is normally distributed is \(\frac{(n - 1)s^{2}}{\chi_{\frac{\alpha}{2},n - 1}^{2}}\leq\sigma^{2}\leq\frac{(n - 1)s^{2}}{\chi_{1-\frac{\alpha}{2},n - 1}^{2}}\), where \(n\) is the sample size, \(s^{2}\) is the sample variance, \(\chi_{\frac{\alpha}{2},n - 1}^{2}\) and \(\chi_{1-\frac{\alpha}{2},n - 1}^{2}\) are the critical values from the chi - square distribution.
Given \(n = 12\), so \(n-1=11\). The confidence level \(C = 0.90\), then \(\alpha=1 - C=0.10\), \(\frac{\alpha}{2}=0.05\), \(1-\frac{\alpha}{2}=0.95\).

Step2: Find the critical values

Using the chi - square distribution table or a statistical software, for \(df=n - 1=11\):
\(\chi_{\frac{\alpha}{2},n - 1}^{2}=\chi_{0.05,11}^{2}=19.675\) and \(\chi_{1-\frac{\alpha}{2},n - 1}^{2}=\chi_{0.95,11}^{2}=4.575\)
The sample standard deviation \(s = 0.89\), so the sample variance \(s^{2}=0.89^{2}=0.792\)

Step3: Calculate the lower and upper bounds

The lower bound:

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The upper bound:

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Answer:

The \(90\%\) confidence interval for the variance of all capsule weights is \((0.443,1.904)\)