Question

A certain chemical pollutant in the Genesee River has been constant for several years with mean μ = 34 ppm (parts per million) and standard deviation σ = 8 ppm. A group of factory representatives whose companies discharge liquids into the river is now claiming that they have lowered the average with improved filtration devices. A group of environmentalists will test to see if this is true at the 4% level of significance. Assume that their sample of size 50 gives a mean of 32.5 ppm.

What is the Z-score? 


What percentage of samples would have a higher ppm of pollutants?

What percentage of samples would have a lower ppm of pollutants?

At a p<.04 level of significance, do you reject or accept the null hypothesis?

What is the critical value for Z at this significance level (see power point)?

Answer 1

Ans:

Z score=(32.5-34)/(8/sqrt(50))

Z score=-1.326

percentage of samples would have a higher ppm of pollutants=100*P(Z>-1.326)=100*0.9076=90.76%

percentage of samples would have a lower ppm of pollutants=100*P(Z<-1.326)=100*0.0924=9.24%

p-value=P(Z<-1.326)=0.0924

As,p-value>0.04,we accept the null hypothesis.

Critical Z value=normsinv(0.04)=-1.75