Question

A civil engineer is analyzing the compressive strength of concrete. A random sample of 12 specimens has a mean compression strength of 3250 psi with variance 1000 (psi)2.
i) Construct a 95% confidence interval for the population mean compressive strength.
ii) Construct a 95% confidence interval for the population standard deviation of compressive strength.

Answer 1

Answer on Question #42056 – Math – Statistics and Probability

A civil engineer is analyzing the compressive strength of concrete. A random sample of 12 specimens has a mean compression strength of 3250 psi with variance 1000 (psi)2.

i) Construct a 95% confidence interval for the population mean compressive strength.

ii) Construct a 95% confidence interval for the population standard deviation of compressive strength

Solution

$$n = 12, \bar{x} = 3250, s = 1000.$$

i)

A 95% confidence interval for the population mean compressive strength is

$$\left(\bar{x} - t_{n-1} \frac{s}{\sqrt{n}}; \bar{x} + t_{\frac{\alpha}{2}, n-1} \frac{s}{\sqrt{n}}\right) = \left(3250 - t_{11} \frac{1000}{\sqrt{12}}; 3250 + t_{11} \frac{1000}{\sqrt{12}}\right).$$

It is $\left(3250 - t_{11} \frac{1000}{\sqrt{12}}; 3250 + t_{11} \frac{1000}{\sqrt{12}}\right) =$

$$\begin{aligned} &= (3250 - \text{TINV}(0,05;11) * 1000 / \text{SQRT}(12); 3250 + \text{TINV}(0,05;11) * 1000 / \text{SQRT}(12)) \\ &= \left(3250 - 2.2 * \frac{1000}{\text{SQRT}(12)}; 3250 + 2.2 * \frac{1000}{\text{SQRT}(12)}\right) = \\ &= \left(3250 - 2.2 * \frac{1000}{3.464}; 3250 + 2.2 * \frac{1000}{3.464}\right) = (2614.630; 3885.370) \end{aligned}$$

using Excel.

Note that Excel gives a two-tailed value for t-value, i.e., we do not divide $\alpha$ in half.

ii)

A 95% confidence interval for the population standard variance of compressive strength is

$$\left(\sqrt{\frac{(n-1)s^2}{\chi_{0.025,11}^2}}; \sqrt{\frac{(n-1)s^2}{\chi_{1-\frac{\alpha}{2},n-1}^2}}\right) = \left(\sqrt{\frac{(12-1)1000^2}{\chi_{0.025,11}^2}}; \sqrt{\frac{(12-1)1000^2}{\chi_{0.975,11}^2}}\right) = \left(\sqrt{\frac{(12-1)1000^2}{21.92}}; \sqrt{\frac{(12-1)1000^2}{3.81}}\right).$$

It is

$$\begin{aligned} &\left(\sqrt{\frac{(12-1)1000^2}{\chi_{0.025,11}^2}}; \sqrt{\frac{(12-1)1000^2}{\chi_{0.975,11}^2}}\right) \\ &= \left(1000 * \text{SQRT}\left(\frac{12 - 1}{\text{CHIINV}(0,025;11)}\right); 1000 * \text{SQRT}\left(\frac{12 - 1}{\text{CHIINV}(0,975;11)}\right)\right) = \\ &= \left(1000 * \sqrt{\frac{12 - 1}{21.92}}; 1000 * \sqrt{\frac{12 - 1}{3.816}}\right) = (708.395; 1697.878) \end{aligned}$$

using Excel.

Answer: i) (2614.630; 3885.370); ii) (708.395; 1697.878).

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