Question

The Environmental Protection Agency sets limits on the maximum allowable concentration of certain chemicals in drinking water. For the substance PCB, the limit was set at 5 ppm (parts per million). The PCB level is unsafe if it is greater than 5 ppm. A random sample of 36 water specimens from the same well results in a mean PCB concentration of 5.2 ppm with standard deviation of 0.6 ppm. Does the data substantiate that the water is unsafe?

Answer 1

Explanation:

To determine if the water is unsafe, we need to perform a hypothesis test. We can use a one-sample t-test with a significance level of 0.05. The null hypothesis is that the true mean PCB concentration is equal to or less than 5 ppm and the alternative hypothesis is that the true mean is greater than 5 ppm.

We can calculate the test statistic using the formula:

t = (sample mean - hypothesized mean) / (standard deviation/sqrt (sample size))

Substituting the values from the problem, we get:

t = (5.2 - 5) / (0.6 / sqrt(36))

t = 2

Using a t-distribution table with 35 degrees of freedom (sample size - 1), we can find the critical value for a one-tailed test with a significance level of 0.05. The critical value is 1.690.

Since our calculated t-value (2) is greater than the critical value (1.690), we reject the null hypothesis. We can conclude that the data provide evidence that the true mean PCB concentration in the water is greater than 5 ppm, and therefore, the water is unsafe.