Question

Certain types of mining result in the release of mildly radioactive by-products, which can get into the water supply. The EPA has issued regulations concerning thits on the radioactivity of municipal drinking water. Specifically, the maximym allowable level is 5 picocuries per liter (pcu/L). A random sample of 24 drinking watefs from a city/s water supply produced a mean of 4.61pcu/L and a standard deviation of 0.87pcu/L. a. Do these data provide evidence that the mean level is safe? Test using α=0.0. b. Explain why you would prefer to use a small value of α for the test in part a.

Answer 1

To determine if the mean level of radioactivity in the drinking water is safe, we can perform a hypothesis test using a one-sample t-test. The sample data does not provide enough evidence to conclude that the mean level is unsafe.

Step-by-step explanation:

To determine if the mean level of radioactivity in the drinking water is safe, we can perform a hypothesis test. The null hypothesis is that the mean level is equal to 5 pcu/L, and the alternative hypothesis is that the mean level is less than 5 pcu/L. We can use a one-sample t-test to test this hypothesis.

To perform the test, we calculate the test statistic t using the formula t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size)). In this case, the sample mean is 4.61 pcu/L, the hypothesized mean is 5 pcu/L, the sample standard deviation is 0.87 pcu/L, and the sample size is 24. Plugging these values into the formula gives us t = (4.61 - 5) / (0.87 / sqrt(24)) = -0.5528.

Next, we determine the critical value for the test. Since the test is one-tailed with a significance level of 0.05, the critical value is -1.711, obtained from a t-distribution table. Since -0.5528 > -1.711, we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that the mean level of radioactivity in the drinking water is unsafe.