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

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, = 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, = 224.6

The sample standard deviation, s₂ = 5.0

The level of significance = 0.10

The null hypothesis, H₀; ≥

The alternative hypothesis, Hₐ;   <

The degrees of freedom = 16 - 1 = 15

The test statistic, = 1.341

Plugging in the values, we get;

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; ≥

y-1700=200(x+4)

step-by-step explanation:

200(x+4) is 200x + 800 she is saving 200 per week x is the weeks and the -1700 is. us that’s howmuch she already had