Answer to Question #109516 in Statistics and Probability for emag

Question

The percentage of titanium in an alloy used in aerospace castings is measured in 51 randomly selected
parts. The sample standard deviation is s = 0.37.
(a) Test the hypothesis H0

: σ = 0.25 versus H1

: σ ≠ 0.25 using α = 0.05. State any necessary

assumptions about the underlying distribution of the data.
(b) Explain how you could answer the question in part (a) by constructing a 95% two-sided
confidence interval for σ^2
.

Answer 1

$a) H_0: \sigma=\sigma_0=0.25, H_1: \sigma\neq\sigma_0=0.25\text{ (two-sided)}.\\ n=51\\ s=0.37\\ \alpha=0.05\\ \text{We will assume that the percantage of titanium has normal}\\ \text{distribution}.\\ \text{We will consider random variable } \chi^2=\frac{(n-1)s^2}{\sigma_0^2}\\ \text{ where }s^2 \text{ is random value of sample variance (corrected).}\\ k=n-1=51-1=50\text{ is degree of freedom}.\\ \chi^2_{l. cr.}=\chi^2_{cr.}(1-\alpha/2;k)=\chi^2_{cr.}(0.975;50)\approx 32.357.\\ \chi^2_{r. cr.}=\chi^2_{cr.}(\alpha/2;k)=\chi^2_{cr.}(0.025;50)\approx 71.42.\\ \text{Critical region:} (-\infty, 32.357)\cup(71.42,\infty).\\ \chi^2=\frac{50\cdot (0.37)^2}{(0.25)^2}=109.52.\\ \text{Our }\chi^2\text{ falls into the critical region. So we reject } H_0.\\ b)\text{Here we have }\alpha =0.95\text{ and we should build two-sided }\\ \alpha\text{-confidence interval for }\sigma^2.\\ \frac{(n-1)s^2}{\chi^2_{\frac{1-\alpha}{2},n-1}} \leq\sigma^2\leq \frac{(n-1)s^2}{\chi^2_{\frac{1+\alpha}{2},n-1}}\\ \frac{(51-1)(0.37)^2}{\chi^2_{0.025,51-1}} \leq\sigma^2\leq \frac{(51-1)(0.37)^2}{\chi^2_{0.975,51-1}}\\ 0.096\leq\sigma^2\leq 0.211\\ \sigma_0=0.25\\ \sigma_0^2=0.0625\\ 0.0625\text{ does not fall into CI. So at a significance level }\alpha=0.05 \text{ we will reject }H_0\text{ (two-sided hypothesis test)}.$