Chapter 8.3 ple of 42 water specimens taken from a construction site, 26 contain

Question

Chapter 8.3 ple of 42 water specimens taken from a construction site, 26 contained Problem 26, page 388: In a sam detectable levels of lead. Construct a 90% confidence interval for of lead. Construct a 95% confidence interval for the proportion of water specimens that contain detectable levels of lead. What is the point-estimate and margin of error? If the manager of the construction company wants a margin of error of +/-03( 396), what size of sample is necessary to achieve this goal for 95% confidence? the proportion of water specimens that contain detectable levels a) b) c)

Explanation / Answer

Solution(a):

Confidence interval can be calculated as

p - Zalpha/2 *sqrt(p(1-p)/n) to  p + Zalpha/2 *sqrt(p(1-p)/n)

where p can be calculated as

p = 26/42 = 0.619048

90% confidence interval can be calculated as follows:

Zalpha/2 = 1.645

0.619 - 1.645*sqrt(0.619048*(1-0.619048)/42) to 0.619 + 1.645*sqrt(0.619048*(1-0.619048)/42)

0.619 - 0.1232 to  0.619 + 0.1232

So 90% confidence interval is

0.742312

Solution(b):

95% confidence interval can be calculated as follows:

p - Zalpha/2 *sqrt(p(1-p)/n) to  p + Zalpha/2 *sqrt(p(1-p)/n)

Zalpha/2 = 1.96

0.619 - 1.96*sqrt(0.619048*(1-0.619048)/42) to 0.619 + 1.96*sqrt(0.619048*(1-0.619048)/42)

Here point estimate is 0.619

Margin of error is 1.96*sqrt(0.619048*(1-0.619048)/42) = 0.146869

So confidence interval is 0.619 - 0.146869 to 0.619 + 0.146869

= 0.472 to 0.765

Solution(c):

Margin of error = Zalpha/2 *sqrt(p(1-p)/n)

Margin of error = 0.03

0.03 = 1.96 * sqrt(0.619*(1-0.619)/n)

0.03 = 1.96 * sqrt(0.2358/n)

n = 1006.4 or 1007

0.495783 to

0.742312

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