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

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