Question

The concentration of hexane (a common solvent) was measured in units of micrograms per liter for a simple random sample of sixteen specimens of untreated ground water taken near a municipal landfill. The sample mean was 228.0 with a sample standard deviation of 4.3. Twenty specimens of treated ground water had an average hexane concentration of 224.6 with a standard deviation of 5.0. It is reasonable to assume that both samples come from populations that are approximately normal.
Can you conclude that the mean hexane concentration is less in treated water than in untreated water? Use the ? = 0.10 level of significance.
No
Yes

Answer 1

Answer:

Yes, it can be concluded that the mean hexane concentration is less in treated water than in unsaturated water

Step-by-step explanation:

The number of of specimen in the samples of untreated water, n₁ = 16

The sample mean, $\overline x_1$ = 228.0

The sample standard deviation, s₁ = 4.3

The number of of specimen in the samples of treated water, n₂ = 20

The sample mean, $\overline x_2$ = 224.6

The sample standard deviation, s₂ = 5.0

The level of significance = 0.10

The null hypothesis, H₀; $\overline x_1$ ≥ $\overline x_2$

The alternative hypothesis, Hₐ;  $\overline x_1$ < $\overline x_2$

The degrees of freedom = 16 - 1 = 15

The test statistic, $t_{\alpha}$ = 1.341

$t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}-\dfrac{s _{2}^{2}}{n_{2}}}}$

Plugging in the values, we get;

$t=\dfrac{(224.6- 228.0)}{\sqrt{\dfrac{5.0^{2}}{20} -\dfrac{4.3^{2} }{16}}} \approx -11.0675$

Given that the t-value is large, the corresponding p-value is low, therefore, we fail to reject the null hypothesis and there is considerable statistical evidence to suggest that the mean hexane concentration is less in treated than in untreated water, therefore, we have; $\overline x_1$ ≥ $\overline x_2$