EPA (Environmental Protection Agency) considers it unsafe if a 100-milliliter sa

Question

EPA (Environmental Protection Agency) considers it unsafe if a 100-milliliter sample (about 3.3 ounces) of water contains more than 400 coliform bacteria. Researchers took water samples at 30 State Park swimming areas in MA. Those samples were taken to laboratories and tested for coliform. An unsafe level of coliform means there's a higher chance that disease-causing bacteria are present and more risk that a swimmer will become ill. Here are the coliform levels found by the laboratories: n = 30, Min = 6, Max = 2503, average = 1105.03, standard deviation = 714.93. Are these data good evidence that, on average, the coliform levels in these swimming areas were unsafe? a. Define mu, H_0 and H_a b. Calculate the test statistic z. c. Find the P-value. d. State your conclusion in context.

Explanation / Answer

Back-up Theory

Let X = Number of coliform bacteria in water sample

We assume X ~ N(µ, 2)

H0: µ = µ0   Vs HA: µ > µ0

Test Statistic: t = (n)(Xbar - µ0)/s

where n = sample size and Xbar = sample mean, s = sample standard deviation

Under H0, t ~ tn - 1

H0 is rejected/accepted at % level of significance if p-value of tcal </>

Given Data and Computations

Given, n = 30, Xbar = 1105.03, s = 714.93, µ0 = 400,

t = (30)(1105.03 - 400)/714.93 = 5.401

p-value = P(t29 > 5.401) = 0.00000416 [using Excel Function], t29 stands for a variable whose distribution is t with degrees of freedom = 29.

Since p-value is very small, even smaller than 0.001 (0.1% level of significance),

H0 is rejected

Now, to work out solution part by part

Answers to all parts of the question are readily available in the above working.

However, for easy reference, the relevant portions are copy-pasted against the parts indicated in the question.

Part (a)

µ0 = 400, H0: µ = µ0 (400)   Vs HA: µ > µ0 (400) ANSWER

Part (b)

Test Statistic = t = (n)(Xbar - µ0)/s

where n = sample size and Xbar = sample mean, s = sample standard deviation

ANSWER

Part (c)

p-value = P(t29 > 5.401) = 0.00000416 ANSWER

Part (d)

Conclusion: H0 is rejected

=> there is strong evidence to suggest that the coliform bacteria in water is well above the permissible limits. ANSWER

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.