Construct a 99% confidence interval for the population variance and standard deviation, given a sample size of 30 and a sample standard deviation of 1.2 mg.

Sample size, $n = 30$
Sample standard deviation, $s = 1.2 \text{ mg}$
Sample variance, $s^2 = (1.2)^2 = 1.44 \text{ mg}^2$
Confidence level = 99%, so $\alpha = 1 - 0.99 = 0.01$
Degrees of freedom, $df = n-1 = 30-1 = 29$

For a 99% confidence interval with $df = 29$:
$\chi^2_{1-\alpha/2, df} = \chi^2_{1-0.01/2, 29} = \chi^2_{0.995, 29} = 13.121$
$\chi^2_{\alpha/2, df} = \chi^2_{0.01/2, 29} = \chi^2_{0.005, 29} = 52.336$

The formula for the confidence interval for population variance is:

Substitute the values:

Rounding to three decimal places:

b) ii) 99% Confidence Interval for Population Standard Deviation ($\sigma$)

The confidence interval for population standard deviation is found by taking the square root of the confidence interval for the population variance:

Rounding to three decimal places:

c) Interpretation of Findings in b(i) and b(ii)

• For the population variance: We are 99% confident that the true population variance of the weights of the allergy medicine is between $0.798 \text{ mg}^2$ and $3.183 \text{ mg}^2$.
• For the population standard deviation: We are 99% confident that the true population standard deviation of the weights of the allergy medicine is between $0.893 \text{ mg}$ and $1.784 \text{ mg}$. This means that if we were to repeat this sampling process many times, 99% of the confidence intervals constructed would contain the true population standard deviation.