In a certain chemical process, it is very important that a particular solution that is to be used as a reactant have a pH of exactly 8.20. A method for determining pH that is available for solutions of this type is known to give measurements that are normally distributed with a mean equal to the actual pH and with a standard deviation of .02. Suppose 10 independent measurements yielded the following pH values:

$\begin{matrix} \text{8.18} & \text{8.17}\\ \text{8.16} & \text{8.15}\\ \text{8.17} & \text{8.21}\\ \text{8.22} & \text{8.16}\\ \text{8.19} & \text{8.18}\\ \end{matrix}$

(a) What conclusion can be drawn at the $\alpha=.10$ level of significance? (b) What about at the $\alpha=.5$ level of significance?

Solution

VerifiedWe are given sample of size $n = 10$. Calculate that the sample mean is

$\begin{align*} \hat{\mu} = \frac{1}{n} \sum_{i = 1}^{10}x_i = 8.179 \end{align*}$

Define hypothesis

$\begin{align*} H_0: \ \mu = 8.2 \\ H_1: \ \mu \neq 8.2 \end{align*}$

where $\mu$ is population mean. We are given that $\sigma = 0.02$ is population standard deviation. We know that test statistics has distribution

$\begin{align*} \sqrt{n} \cdot \frac{\bar{X} - \mu}{\sigma} \Big|_{H_0} \sim AN(0,1) \end{align*}$

The $p$-value of this random sample is

$\begin{align*} p\text{-value} &= P_{H_0} \left( \Big| \sqrt{n} \cdot \frac{\bar{X} - \mu}{\sigma} \Big| > \Big| \sqrt{n} \cdot \frac{\hat{\mu} - \mu}{\sigma} \Big| \right) \\ &= P_{H_0} \left( \Big| \sqrt{n} \cdot \frac{\bar{X} - \mu}{\sigma} \Big| > \Big| \sqrt{10} \cdot \frac{8.179- 8.2}{0.02} \Big| \right) \\ &= P_{H_0} \left( \Big| \sqrt{n} \cdot \frac{\bar{X} - \mu}{\sigma} \Big| > 3.32 \right) = 2(1 - \Phi(3.32)) \\ &= 0.001 \end{align*}$

where $\Phi$ is CDF of standard normal and we have used table of CDF of standard normal in appendix to obtain final answer. Since $p$-value is less than significance levels in (a) and (b), we reject the null hypothesis in both cases.