Identify K the constant of variation, for each direcr variation. -2y=5x
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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, = 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; ≥
