Question

Unfortunately, arsenic occurs naturally in some groundwater. A mean arsenic level of \mu=8.0 parts per billion (ppb) is considered safe for agricultural use. A well in Texas us used to water cotton crops. This well is tested regular basis for arsenic. A random sample of 37 tests gave a sample mean \bar{x}=7.2 pbb of arsenic, with s = 1.9 pbb. Does this information indicate that the mean of arsenic in this well is less than 8 pbb? Use \alpha=0.01

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq 8$

$H_1:\mu<8$

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ $df=n-1=36$ degrees of freedom, and the critical value for a left-tailed test is $t_c = -2.434494.$

The rejection region for this left-tailed test is $R = \{t: t < -2.434494\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{X}-\mu}{s/\sqrt{n}}=\dfrac{7.2-8}{1.9/\sqrt{37}}\approx -2.561163$$

Since it is observed that $t = -2.561163 <-2.434494= t_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for left-tailed, $df=36$ degrees of freedom, $t=-2.561163$ is $p=0.007384,$ and since $p=0.007384<0.01=\alpha,$  it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is less than $8,$ at the $\alpha = 0.01$significance level.