Question 5 of 8

Question 5 (20 marks: (a) 7, (b) 7, (c) 6)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) 99% CI for μ and for σ. (b) Test $H_0: \mu=8.4$ at α=0.05. (c) Test $H_0: \sigma=0.4$ at α=0.05.

Approach. Compute $\bar{x}$ and $s^2$ from the sums, then use the t-distribution (df=15) for the mean CI and mean test, and the chi-square distribution (df=15) for the variance CI and variance test, since X is Normal and n is small.

1. Preliminary: sample mean and variance. $$\bar{x} = \frac{136.00}{16} = 8.5, \qquad s^2 = \frac{\sum X^2 - n\bar{x}^2}{n-1} = \frac{1159.75-16(8.5)^2}{15} = \frac{3.75}{15} = 0.25, \qquad s = \boxed{0.5}.$$
2. (a)(i) 99% CI for the mean, df = 15. $t_{0.005,15}=2.947$: $$\bar{x} \pm t_{0.005,15}\frac{s}{\sqrt{n}} = 8.5 \pm 2.947\left(\frac{0.5}{4}\right) = 8.5\pm 0.368, \qquad \boxed{(8.132,\ 8.868)}.$$
3. (a)(ii) 99% CI for the standard deviation. With $\chi^2_{0.995,15}=4.601$ and $\chi^2_{0.005,15}=32.801$: $$\sqrt{\frac{(n-1)s^2}{\chi^2_{0.005,15}}} \le \sigma \le \sqrt{\frac{(n-1)s^2}{\chi^2_{0.995,15}}}, \qquad \boxed{(0.338,\ 0.903)}.$$
4. (b) Test H₀: μ = 8.4 vs H₁: μ ≠ 8.4. $$t = \frac{\bar{x}-8.4}{s/\sqrt{n}} = \frac{8.5-8.4}{0.5/4} = 0.80.$$ Since $t_{0.025,15}=2.131$ and $|0.80| < 2.131$, we fail to reject H₀ — the mean is not significantly different from 8.4.
5. (c) Test H₀: σ = 0.4 vs H₁: σ ≠ 0.4. $$\chi^2 = \frac{(n-1)s^2}{\sigma_0^2} = \frac{15(0.25)}{0.16} = 23.44.$$ Since $\chi^2_{0.975,15}=6.262$ and $\chi^2_{0.025,15}=27.488$, and $6.262<23.44<27.488$, we fail to reject H₀ — the standard deviation is not significantly different from 0.4.