Question

Contaminated water: In a sample of

42

water specimens taken from a construction site,

26

contained detectable levels of lead.

Part 1 of 3

Your Answer is correct

(a) Construct a

90%

confidence interval for the proportion of water specimens that contain detectable levels of lead. Round the answer to at least three decimal places.

A

90%

confidence interval for the proportion of water specimens that contain detectable levels of lead is

0.497<<p0.743

.

Part: 1 / 3

1 of 3 Parts Complete

Part 2 of 3

(b) Construct a

99.8%

confidence interval for the proportion of water specimens that contain detectable levels of lead. Round the answer to at least three decimal places.

A 99.8% confidence interval for the proportion of water specimens that contain detectable levels of lead is <<p .

Answer 1

Solution :

Given that,

n = 42

x = 26

Point estimate = sample proportion = = x / n = 26 /42 = 0.619

1 - = 1-0.619 = 0.381

At 90% confidence level

= 1-0.90% =1-0.90 =0.10

/2 =0.10/ 2= 0.05

Z/2 = Z0.05 = 1.645

Z/2 = 1.645  

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 * ((0.619*(0.381) /42 )

= 0.1233

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.619 -0.1233< p < 0.619 +0.1233

0.497 < p < 0.743

(0.497 ,0743 )

90% confidence interval for the proportion of water specimens that contain detectable levels of lead is

0.497 and 0.743.

2)

n = 42

x = 26

Point estimate = sample proportion = = x / n = 26 /42 = 0.619

1 - = 1-0.619 = 0.381

Z/2 = 3.090

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 3.090 * ((0.619*(0.381) /42 )

= 0.232

A 99.8% confidence interval for population proportion p is ,

- E < p < + E

0.619 -0.232 < p < 0.619 +0.232

0.387< p < 0. 851

( 0.387 , 0.851)

99.8% confidence interval for the proportion of water specimens that contain detectable levels of lead is

0.387 and 0.851.