The Virginia Department of Environmental Quality (VDEQ) uses probabilistic monitoring to regulate the water quality of streams in the Commonwealth of Virginia. Of the 85 Eastern Virginia Sites (group 1), 24 do not meet minimum requirements. Of the 80 units sampled in Western Virginia Sites (group 2), 17 do not meet minimum requirements. Assume the data can be treated as independent simple random samples. To determine whether the proportions of streams that fail to meet minimum requirements in the two areas are equal or not, test the following hypotheses.
a) State the null and alternative hypotheses in order to determine whether the proportions of streams that fail to meet minimum requirements in the two areas are equal or not .
(b) Calculate the value of the test statistic z.
(c) Determine the P-value associated with this observed value of z and draw the conclusion about the hypotheses using alpha=0.05.

c) Since our p-value would be 0.29834, this is higher than our significance level of a=0.05, so we fail to reject the null hypothesis and conclude that there is sufficient evidence to prove that the two proportions are equal.

shubhamchouhanvrVT shubhamchouhanvrVT · Answer 2 · 2022-05-14T15:18:00+02:00

a) The null hypothesis is given by Hn: p1=p2, Ha: p1≠p2

What is null hypothesis?

The hypothesis that there is no significant difference between specified populations, any observed difference being due to sampling or experimental error.

Given that:

p1=24/85=0.2824

p2=17/80=0.2125

p=(24+17)/(85+80)=41/165=0.2485

n1=85

n2=80

Z=(p1-p2)/[p(1-p)(1/n1+1/n2)

Z=(0.2824-0.2125)/[0.2485(1-0.2485)(1/85+1/80)

Z=1.0384

c) Since our p-value would be 0.29834, this is higher than our significance level of a=0.05, so we fail to reject the null hypothesis and conclude that there is sufficient evidence to prove that the two proportions are equal.

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