A Lab has been asked to evaluate the claim that drinking water in a local restaurant has a lead concentration of 6 parts per billion (ppb). Repeated measurements follow a normal distribution and the population standard deviation is taken to be 0.25 ppb. α = 0.01.

• A sample of three measurements is taken and finds: 6.79; 6.13; 7.17

  1. Is there evidence to suggest that the lead concentration is different from 6 ppb?

This is a homework question that I'm really stuck on, so if someone could point me in the right direction that would be great.

I realise that it is a hypothesis testing question and that the null hypothesis is that µ = 6 ppb while the alternative is that µ does not equal 6 ppb. However, I really don't know what to do with the three sample measurements. Do I get the mean of the three figures or do I use each in a separate hypothesis test?

  • 1
    $\begingroup$ This is a one-sample Z test. You do indeed do calculations based on the sample mean. example $\endgroup$ – Glen_b Apr 29 '13 at 23:46

From a frequentist perspective you should calculate the standard error of the sample mean (using the known population standard deviation). You can then work out how many standard errors the sample mean is away from six ppb and look at the corresponding Normal probability density function tails (note the plural).

The assumption is that there's a fixed lead concentration in the water & the measurement error is what's random - distributed normally with mean zero, i.e. there's no systematic error in the measurements.


Here's the Bayesian answer. How would the Bayesian estimate the population mean? Assuming a very noninformative prior (with precision zero, see comments), that's simply the average: $$\mu'=1/n\sum_{i=1}^n x_i\approx 6.7$$ Since you know the population standard deviation, you know the population variance $Var=SD^2$, and you know the precision $\rho=1/Var=16$. The Bayesian learning rule for the precision is the sum of the precisions of your observations. You have 3, so $\rho'=3*16=48$, $Var=1/48$, $SD=\sqrt{1/48}\approx 0.144$. So the Bayesian thinks the population mean is normally distributed with mean $\mu'$ and SD $0.144$.

The question is: Given your estimate from the observations, how likely is it that the test value 6ppb or less can originate from $N(\mu',\rho')$? We check a standard normal table. Computing the Z score: $$Z=(6-\mu')/sd=-0.7/0.144\approx -4.86.$$ The cumulative table actually only goes as far as $Z=3$, at which point $0.9987$ of the values are below $Z$, hence $1-0.9987$ of the values are below $Z=-3$ (by symmetry of the normal) and even fewer below $Z=-4.8$. Thus, you can reject the hypothesis that the estimated population mean is equal to the test value $6$ with your $\alpha$ level, since $1-0.9987<\alpha$.

  • $\begingroup$ Why would a Bayesian necessarily estimate the population mean as the sample average? $\endgroup$ – Scortchi Apr 29 '13 at 22:34
  • $\begingroup$ More generally, the Bayesian estimate is a precision weighted average. Since the precision/sd of every observation is the same, this is identical to the sample average (in this case). $\endgroup$ – Nameless Apr 29 '13 at 22:49
  • $\begingroup$ What about the prior? $\endgroup$ – Scortchi Apr 29 '13 at 23:15
  • $\begingroup$ Hehe you're right. For the above to be correct we could assume the prior is normal with infinite variance or zero precision. Then it doesn't show up. $\endgroup$ – Nameless Apr 29 '13 at 23:22
  • $\begingroup$ Worth putting that in the answer. $\endgroup$ – Scortchi Apr 29 '13 at 23:25

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