Question 5 of 8: Confidence Intervals and Hypothesis Tests on Bicarbonate Content

Question 5: Confidence Intervals and Hypothesis Tests on Bicarbonate Content (20 marks)

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. $n=14$, $\sum X=4{,}956.0$ mg/L, $\sum X^2=1{,}754{,}476.0$ (mg/L)$^2$; $X$ assumed Normal, $\sigma$ unknown (estimated by $s$).

Find. (a) 99% CI for $\mu$ and for $\sigma$; (b) test $H_0:\mu=350$ vs. $H_1:\mu\ne350$ at $\alpha=0.05$; (c) test $H_0:\sigma=3.2$ vs. $H_1:\sigma\ne3.2$ at $\alpha=0.05$.

Approach. Reduce the sums to $\bar x$ and $s^2$ first; use the $t$-distribution for the mean (small $n$, $\sigma$ unknown) and the chi-square distribution for the variance/sd.

1. Summary statistics. $\bar x=\dfrac{\sum X}{n}=\dfrac{4{,}956.0}{14}=354.0\text{ mg/L}$. $$s^2=\dfrac{\sum X^2-(\sum X)^2/n}{n-1}=\dfrac{1{,}754{,}476.0-4{,}956.0^2/14}{13}=\dfrac{52.0}{13}=\boxed{4.0}\ (\text{mg/L})^2,\quad s=2.0\text{ mg/L}$$
2. (a)(i) 99% CI for $\mu$. With $n-1=13$ df and $\alpha=0.01$, $t_{0.005,13}=3.012$. $$\bar x\pm t_{0.005,13}\dfrac{s}{\sqrt n}=354.0\pm(3.012)\dfrac{2.0}{\sqrt{14}}=354.0\pm1.610\Rightarrow\boxed{(352.39,\ 355.61)\text{ mg/L}}$$
3. (a)(ii) 99% CI for $\sigma$. Using $\chi^2_{0.995,13}=3.565$ and $\chi^2_{0.005,13}=29.82$ (from the chi-square table), $$\dfrac{(n-1)s^2}{\chi^2_{0.005,13}}\le\sigma^2\le\dfrac{(n-1)s^2}{\chi^2_{0.995,13}}\ \Rightarrow\ \dfrac{52.0}{29.82}\le\sigma^2\le\dfrac{52.0}{3.565}\ \Rightarrow\ 1.744\le\sigma^2\le14.59$$ Taking square roots, $$\boxed{1.32\text{ mg/L}\le\sigma\le3.82\text{ mg/L}}$$
4. (b) Test $H_0:\mu=350$ vs. $H_1:\mu\ne350$. Test statistic $t=\dfrac{\bar x-\mu_0}{s/\sqrt n}=\dfrac{354.0-350.0}{2.0/\sqrt{14}}=\dfrac{4.0}{0.5348}=\boxed{7.483}$. Since $|7.483|\gg t_{0.025,13}=2.160$: reject $H_0$ — the true mean bicarbonate content IS significantly different from (above) 350.0 mg/L.
5. (c) Test $H_0:\sigma=3.2$ vs. $H_1:\sigma\ne3.2$. Test statistic $\chi^2=\dfrac{(n-1)s^2}{\sigma_0^2}=\dfrac{13(4.0)}{3.2^2}=\dfrac{52.0}{10.24}=\boxed{5.078}$. The two-sided acceptance region at $\alpha=0.05$, 13 df is $\big(\chi^2_{0.975,13},\,\chi^2_{0.025,13}\big)=(5.009,\,24.74)$. Since $5.009\lt5.078\lt24.74$: fail to reject $H_0$ — the true standard deviation is not significantly different from 3.2 mg/L.