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 found an average chemical waste of 187 g per mile with a standard deviation of 14.54 g. Round to the margin of error and standard error to the 5th decimal place

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

(b) Find the 95% confidence interval (in g per mile) for this study. (Use a table or technology. 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. It would stay the same. It would go to zero?

d) What would happen to the margin of error if the confidence level increased? It would also increase, It would decrease, It would stay the same, It would go to zero?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 187

sample standard deviation = s = 14.54

sample size = n = 20

a) Degrees of freedom = df = n - 1 = 20-1 =19

t _/2,df = 2.09

Standard error =(s /n) =( 14.54/ 20) = 3.25124

Margin of error = E = t_/2,df * (s /n)

= 2.09 * (14.54 / 20)

Margin of error = E = 6.79509

The 95% confidence interval estimate of the population mean is,

- E < < + E

187- 6.79509 < < 187+ 6.79509

180.205 < < 193.795

(180.208 g ,193.795 g )

c)  The margin of error if the sample size increased =

Answer = It would decrease

d)  The margin of error if the confidence level increased

Answer = It would also increase