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By law, an industrial plant can discharge no more than 500 gallons of waste water per hour, on the average, into a neighboring lake. Based on other infractions they have noticed, an environmental action group believes this limit is being exceeded. Monitoring the plant is expensive, and a random sample of four hours is selected over a period of a week. Software reports:
Variable No. Cases Mean SD SE of Mean
WASTE 4 1000.0 400.0 200.0
(a) Test whether the mean discharge equals 500 gallons per hour against the alternative that the limit is being exceeded. Find the P-valuc and interpret.
(b) Explain why the test may be highly approximate or even invalid if the population distribution of discharge is far from normal.
(c) Explain how your one-sided analysisimplicitly tests the broader null hypothesis that µ <>

Sagot :

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Answer:

P = 0.0438 ; We reject the Null and conclude that there is significant evidence that the discharge limit per gallon of waste has been exceeded.

Due to the small sample size.

Step-by-step explanation:

H0: μ = 500

H1 : μ > 500

Test statistic :

(xbar - μ) / S.E

Tstatistic = (1000 - 500) / 200

= 500 / 200

= 2.5

Pvalue from Test score ; df = 4 - 1 = 3 ; using calculator :

Pvalue = 0.0438

At α = 0.05

Pvalue < α ; We reject the Null and conclude that there is significant evidence that the discharge limit per gallon of waste has been exceeded.

The test above has a very small sample size, and for a distribution to be approximately Normal, the sample size must be sufficiently large enough according to the Central limit theorem.

For a two sided analysis ; the Pvalue is twice that for the one sided, hence, Pvalue = (0.0438 * 2) = 0.0876 yielding a less strong evidence against the Null.

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