Question

The amount of lateral expansion (mils) was determined for a sample of n = 9 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s = 2.84 mils. Assuming normality, derive a 95% CI for ?2 and for ?. (Round your answers to two decimal places.)

Answer 1

Concepts and reason

Confidence interval:

A range of values such that the population parameter can be expected to contain for the given confidence level is termed as the confidence interval. In other words, it can be defined as an interval estimate of the population parameter which is calculated for the given data based on a point estimate and for the given confidence level.

Moreover, the confidence level indicates the possibility that the confidence interval can contain the population parameter. Usually, the confidence level is denoted by . The value is chosen by the researcher. Some of the most common confidence levels are 90%, 95%, and 99%.

Fundamentals

A $100\left( {1 - \alpha } \right)\%$ confidence interval on ${\sigma ^2}$ is

$\frac{{\left( {n - 1} \right){s^2}}}{{\chi _{\alpha /2,n - 1}^2}} \le {\sigma ^2} \le \frac{{\left( {n - 1} \right){s^2}}}{{\chi _{1 - \alpha /2,n - 1}^2}}$ $.....\left( I \right)$

A $100\left( {1 - \alpha } \right)\%$ confidence interval on $\sigma$ is

$\sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _{\alpha /2,n - 1}^2}}} \le \sigma \le \sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _{1 - \alpha /2,n - 1}^2}}}$ $.....\left( {II} \right)$

Here, $\chi _{\alpha /2,n - 1}^2$ and $\chi _{1 - \alpha /2,n - 1}^2$ are the upper and lower $100\alpha /2$ percentage points of the chi-square distribution with $n - 1$ degrees of freedom, respectively.

Where, n is the sample size, ${s^2}$ is the sample variance, $\chi _{\frac{\alpha }{2}}^2$ is the critical value for lower limit and $\chi _{1 - \frac{\alpha }{2}}^2$ is the critical value for upper limit.

Given information:

Sample size, $n = 9$

Sample standard deviation, $s = 2.84$

Confidence level, $1 - \alpha = 0.95$

Implies, level of significance, $\alpha = 0.05$

Degrees of freedom,

$\begin{array}{c}\\df = n - 1\\\\ = 9 - 1\\\\ = 8\\\end{array}$

At $\alpha = 0.05$ with $df = 8$ , the critical values from the chi-square table:

$\begin{array}{l}\\\chi _{\alpha /2,n - 1}^2 = \chi _{0.025,8}^2 = 17.535\\\\\chi _{1 - \alpha /2,n - 1}^2 = \chi _{0.975,8}^2 = 2.180\\\end{array}$

Substitute the above values in equation (I), you will get the 95% confidence interval for ${\sigma ^2}$ .

$\begin{array}{c}\\\frac{{\left( {n - 1} \right){s^2}}}{{\chi _{\alpha /2,n - 1}^2}} \le {\sigma ^2} \le \frac{{\left( {n - 1} \right){s^2}}}{{\chi _{1 - \alpha /2,n - 1}^2}}\\\\\frac{{\left( {9 - 1} \right){{\left( {2.84} \right)}^2}}}{{17.535}} \le {\sigma ^2} \le \frac{{\left( {9 - 1} \right){{\left( {2.84} \right)}^2}}}{{2.180}}\\\\3.679772 \le {\sigma ^2} \le 29.59853\\\\3.680 \le {\sigma ^2} \le 29.599\\\end{array}$

Given information:

Sample size, $n = 9$

Sample standard deviation, $s = 2.84$

Confidence level, $1 - \alpha = 0.95$

Implies, level of significance, $\alpha = 0.05$

Degrees of freedom,

$\begin{array}{c}\\df = n - 1\\\\ = 9 - 1\\\\ = 8\\\end{array}$

At $\alpha = 0.05$ with $df = 8$ , the critical values from the chi-square table:

$\begin{array}{l}\\\chi _{\alpha /2,n - 1}^2 = \chi _{0.025,8}^2 = 17.535\\\\\chi _{1 - \alpha /2,n - 1}^2 = \chi _{0.975,8}^2 = 2.180\\\end{array}$

Substitute the above values in equation (II), you will get the 95% confidence interval for $\sigma$ .

$\begin{array}{c}\\\frac{{\left( {n - 1} \right){s^2}}}{{\chi _{\alpha /2,n - 1}^2}} \le \sigma \le \frac{{\left( {n - 1} \right){s^2}}}{{\chi _{1 - \alpha /2,n - 1}^2}}\\\\\sqrt {\frac{{\left( {9 - 1} \right){{\left( {2.84} \right)}^2}}}{{17.535}}} \le \sigma \le \sqrt {\frac{{\left( {9 - 1} \right){{\left( {2.84} \right)}^2}}}{{2.180}}} \\\\\sqrt {3.679772} \le \sigma \le \sqrt {29.59853} \\\\1.918 \le \sigma \le 5.440\\\end{array}$

Ans:

The 95% confidence interval for the variance of amount of lateral expansion (mils) is (3.680, 29.599).

The 95% confidence interval for the standard deviation of amount of lateral expansion (mils) is (1.918, 5.440).