Question

1. [30] Let X denote the concentration of calcium carbonate (CaCO3) in milligrauns per liter. Following are 10 observations of X: 132.7 134.8 128.8 129.9 131.5 131.7 132.7 131.5 131.2 120.8 (a) Find a 99% confidence interval for the mean concentration of calcium carbonate assuming that the variance of the concentration is known to be σ2-9 ANSWER (b) Find a 99% confidence interval for the mean concentration of calcium carbonate assuming that the variance of the concentration is not known. ANSWER (c) Assume again that the variance of the concentration is not known. Read Exercise 7.1-16 from the text and then find a 99% confidence interval for the variance σ2 of the concentration of calcium carbonate. Show your work in details.

Answer 1

Sample size, n = 10

Mean of the sample =

$\bar x =\frac{\Sigma x_i}{n} = \frac{1305.6}{10} =130.56$

Sample variance =

$s^2=\frac{\Sigma (x_i-\bar x)^2 }{n-1} =14.378$

a) Known Population Variance, $\small \sigma^2$ = 9

At $\alpha$ = 0.01, two tailed critical value, $\small z_{\alpha/2}$ = 2.576

99% confidence interval for the mean:

$\small \left (\bar x \pm z_{\alpha/2}*\sqrt{\frac{\sigma^2}{n}} \right )$

$\small \left (130.56 \pm 2.576*\sqrt{\frac{9}{10}} \right )$

$\small \left (\textbf{128.116, 133.004} \right )$

b) Unknown Population Variance

At $\alpha$ = 0.01 and df = 10-1= 9, two tailed critical value, $\small t_{\alpha/2}$ = 3.25

99% confidence interval for the mean:

$\small \left (\bar x \pm t_{\alpha/2}*\sqrt{\frac{s^2}{n}} \right )$

$\small \left (130.56 \pm 3.25*\sqrt{\frac{14.378}{10}} \right )$

$\small \left (\textbf{126.663, 134.457} \right )$

c) At $\small \alpha$ = 0.01, critical value are, $\small \chi^2_{\alpha /2} = 23.589 \textup{ and }\chi^2_{1-\alpha /2} = 1.735$

99% confidence interval for population variance:

$\small \left (\frac{(n-1)s^2}{\chi ^2_{\alpha/2}}<\sigma ^2<\frac{(n-1)s^2}{\chi ^2_{(1-\alpha/2)}} \right )$

$\small \left (\frac{(10-1)*14.378}{23.589}<\sigma ^2<\frac{(10-1)*14.378}{1.735} \right )$

$\small \left (\textbf{5.486}<\sigma ^2<\textbf{74.583} \right )$