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1. The department of nature resources reports that a fish is unsafe to eat if the polychlorinated biphenol (PCB) concentration exceeds 4.0 parts billion (ppb). 10 fish were taken randomly from a local lake and the value of the PCB concentration for each fish was measured. The data are listed blow. 2.9 7.6 4.8 5.2 5.1 4.7 6.9 4.9 3.7 3.8 Let µ = the true mean value of the PCB concentration for all the fish in the lake. We want to test to determine whether the fish from this lake should not be eaten. Write down the null and alternative hypotheses

Sagot :

akiran007

Answer:

H0 rejected as μ>4

and a fish is unsafe to eat from the lake.

Step-by-step explanation:

1)Let the null hypothesis be  H0 : mu ≤4 against the alternate Ha: mu > 4 ppb

H0: μ ≤4   against the claim Ha:  μ>4

2) X`= ∑x/ n= 2.9 +7.6+ 4.8+ 5.2+ 5.1+ 4.7+ 6.9+ 4.9+ 3.7+ 3.8/10

= 49.6/10= 4.96

The standard deviation can be calculated  sigma= 1.344767 ( using statistic calculator)

3) The significance level is taken to be  ∝=0.05

The value of z at 0.05 for 1 sided test is z >± 1.645

i.e the critical region is less than - 1.645 and greater than +1.645

4) Taking the distribution to be approximately normal

Z= x`- u / s/ √n

Z= 4.96-4/ 1.345/10

Z= 4.96-4/ 1.345/3.1622

Z= 0.96/0.4253

Z= 2.257

5) Since the calculated value of z = 2.257 falls in the critical region we reject our null hypothesis and conclude that true mean value of the PCB concentration is greater than 4 (ppb) and that a fish is unsafe to eat from the lake.

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