In a study of contamination of fish in the nations' rivers and lakes, the EPA found that 91% of water quality test sites showed the presence of PCB, a cancer-causing agent. Suppose a follow-up study of 200 rivers and lakes in 2018 showed the presence of PCB in 177 cases. Does the statistical evidence support the conclusion that as of 2018 water clean-up programs have reduced the proportion of locations with PCB? Use as level of significance. Based on the above, the hypothesis statement and decision are A. The alternative hypothesis is less than.91/Do not reject the null B. The null hypothesis is greater than and equal to.886/Do not reject the null C. The null hypothesis is less than .855/Do not accept the nul D. The alternative hypothesis is greater than .91 /Do not acoept the null E. The nullis less than .91 /Do not reject the null

A. The alternative hypothesis is less than.91/Do not reject the null

Null hypothesis: H0 = 0.91

Alternative hypothesis: Ha < 0.91

z = -1.24

P value = P(Z<-1.24) = 0.11

Rule

If;

P-value > significance level --- accept Null hypothesis

P-value < significance level --- reject Null hypothesis

Z score > Z(at 95% confidence interval) reject Null hypothesis

Z score < Z(at 95% confidence interval) accept Null hypothesis

Step-by-step explanation:

Given;

n=200 represent the random sample taken

Null hypothesis: H0 = 0.91

Alternative hypothesis: Ha < 0.91

Test statistic z score can be calculated with the formula below;

z = (p^−po)/√{po(1−po)/n}

Where,

z= Test statistics

n = Sample size = 200

po = Null hypothesized value = 0.91

p^ = Observed proportion = 177/200 = 0.885

Substituting the values we have

z = (0.885-0.91)/√(0.91(1-0.91)/200)

z = −1.23541552776

z = -1.24

To determine the p value (test statistic) at 0.05 significance level, using a one tailed hypothesis.

P value = P(Z<-1.24) = 0.107488 = 0.11

Since z at 0.05 significance level is between -1.96 and +1.96 and the z score for the test (z = -1.24) which falls with the region bounded by Z at 0.05 significance level. And also the one-tailed hypothesis P-value is 0.11 which is higher than 0.05. Then we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that at 5% significance level the null hypothesis is valid, therefore do not reject null.

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step-by-step explanation:

divide, then simplify your answer.

1. make -15.66 ÷   5.8 a fraction. then divide.

2. simplify.  

3. final answer.