Question

- Public health and safety officials have classified the Zika virus as a hazardous threat to pregnant women in Brazil. A study based in Rio found the virus to be more active near a body of water that was found to be polluted. A study of 20 water samples around areas that were heavily affected by the Zika virus found an average chemical waste of 239 g per mile with a standard deviation of 15.86 g.

(a)What are the degrees of freedom for this study?

______

(b) Construct a 95% confidence interval (in g per mile) for the mean chemical waste of bodies of water for all areas where the Zika virus was overly active. (Round your answers to three decimal places.)

(_____, _____)

g per mile

(c) What would happen to the margin of error if the sample size increased?

It would also increase.It would decrease.

or

It would stay the same.It would go to zero.

(d) Interpret the confidence interval?

a. We are 95% confident that the mean chemical waste of bodies of water for all areas where the Zika virus was overly active is between 231.577 and 246.423.

b. We are 95% confident that 95% of all areas pollution level is between 231.577 and 246.423.

c. We are 5% confident that the mean chemical waste of bodies of water for all areas where the Zika virus was overly active is between 231.577 and 246.423.95% of all areas pollution levels is between 231.577 and 246.423.

Answer 1

(A) sample size is n = 20

degree of freedom = n-1

= 20-1

= 19

(B) t critical value using t distribution table with degree of freedom 19 is 2.09

mean is 239 and standard deviation is 15.86, sample size is n = 20

confidence interval =

(C) Margin of error will decrease because the sample size is inversely proportional to the margin of error.

(D) a. We are 95% confident that the mean chemical waste of bodies of water for all areas where the Zika virus was overly active is between 231.577 and 246.423.