Question

Fecal pollution at Huntington Beach. California mandates fecal indicator bacteria monitoring at all public beaches. When the concentration of fecal bacteria in the water exceeds a certain limit (400 colony-forming units of fecal coliform per 100 millilitres), local health officials must post a sign (called surf zone posting) warning beachgoers of potential health risks. For fecal bacteria, the state uses a single-sample standard; if the fecal limit is exceeded in a single sample of water, surf zone posting is mandatory. This single-sample standard policy has led to a recent rash of beach closures in California. A study of the surf water quality at Huntington Beach in California was published in Environmental Science & Technology (September 2004). The researchers found that beach closings were occurring despite low pollution levels in some instances, while in others, signs were not posted when the fecal limit was exceeded. They attributed these "surf zone posting errors" to the variable nature of water quality in the surf zone (for example, fecal bacteria concentration tends to be higher during ebb tide and at night) and the inherent time delay between when a water sample is collected and when a sign is posted or removed. To prevent posting errors, the researchers recommend using an averaging method rather than a single sample to determine unsafe water quality. (For example, one simple averaging method is to take a random sample of multiple water specimens and compare the average fecal bacteria level of the sample with the limit of 400 CFU/100 mL to determine whether the water is safe.) Discuss the pros and cons of using the single sample standard versus the averaging method. Part of your discussion should address the probability of posting a sign when the water is safe and the probability of posting a sign when the water is unsafe. (Assume that the fecal bacteria concentrations of water specimens at Huntington Beach follow an approximately normal distribution.

Short answer

It can be concluded that by taking more samples, we will get correct decisions rather than taking a single sample.

Step-by-step solution

Given information

The fecal bacteria concentration of water specimens follows a normal distribution with a mean of 370 and a standard deviation of 30.

Calculating the probabilities

Let us assume that one sample is selected.

$\begin{array}{c}P(x\ge 400)=P\left(\frac{\overline{x}-\mu }{\sigma }\ge \frac{400-\mu }{\sigma }\right)\\ =\left(z\ge \frac{400-370}{30}\right)\\ =(z\ge \frac{30}{30})\end{array}$

$=\mathrm{P}(\mathrm{z}\ge 1)$

From the z-score table,

$\begin{array}{l}=0.5-0.3413\\ =0.1587\end{array}$

Hence, it can be concluded that the beach will be closed 15.87% of the time if the water is safe.

Let,

$\begin{array}{c}\mathrm{P}(\mathrm{x}\le 400)=\mathrm{P}\left(\frac{\overline{\mathrm{x}}-\mathrm{\mu }}{\mathrm{\sigma }}\le \frac{400-\mathrm{\mu }}{\mathrm{\sigma }}\right)\\ =\mathrm{P}\left(\mathrm{z}\le \frac{400-430}{30}\right)\\ =\mathrm{P}(\mathrm{z}\le -1)\end{array}$

From the z-score table,

$\begin{array}{l}=0.5-0.3413\\ =0.1587\end{array}$

Hence, it can be concluded that the beach will be open 15.87% of the time.

Let's assume that the fecal bacteria concentration of water specimens follows a normal distribution with a mean of 370 and a standard deviation of 30.

Let's assume that 36 samples are selected.

According to the Central limit theorem, if n is larger, then $\overline{x}$ it follows a normal distribution with mean ${\mu }_{\overline{x}}=\mu$and standard deviation.

$\begin{array}{c}{\mathrm{\sigma }}_{\overline{\mathrm{x}}}=\frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}\\ =\frac{30}{\sqrt{36}}\\ =5\end{array}$

Now, let's,

$\begin{array}{c}\mathrm{P}(\mathrm{x}\ge 400)=\mathrm{P}\left(\frac{\overline{\mathrm{x}}-\mathrm{\mu }}{\frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}}\ge \frac{400-\mathrm{\mu }}{\frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}}\right)\\ =\mathrm{P}\left(\mathrm{z}\ge \frac{400-370}{5}\right)\\ =\mathrm{P}\left(\mathrm{z}\ge \frac{30}{5}\right)\end{array}$

$=P(z\ge 6)$

From the z-score table,

$\begin{array}{l}=0.5-0.5\\ =0\end{array}$

Hence, for the sample size of 36, it can be concluded that the beach will never be closed if the water is Safe.

$\begin{array}{c}\mathrm{P}(\mathrm{x}\le 400)=\mathrm{P}\left(\frac{\overline{\mathrm{x}}-\mathrm{\mu }}{\frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}}\le \frac{400-\mathrm{\mu }}{\frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}}\right)\\ =\mathrm{P}\left(\mathrm{z}\le \frac{400-370}{5}\right)\\ =\mathrm{P}\left(\mathrm{z}\le \frac{30}{5}\right)\end{array}$

$=\mathrm{P}(\mathrm{z}\le -6)$

From the z-score table,

$\begin{array}{l}=0.5-0.5\\ =0\end{array}$

Hence, for the sample size of 36, it can be concluded that the beach will never be open if the water is unsafe.

Thus, it can be concluded that by taking more samples, we will get correct decisions rather than taking a single sample.