Question

In a sample of 59 water specimens taken from a construction site, 22 contained detectable levels of lead. What is the upper bound for the 90% confidence interval for the proportion of water specimens that contain detectable levels of lead. Round to three decimal places (for example: 0.419). Write only a number as your answer.

Answer 1

Solution :

Given that,

n = 59

x = 22

Point estimate = sample proportion = = x / n = 0.373

1 - =0.627

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z _0.05 = 1.645

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

= 1.645 * (((0.373*0.627) / 59)

= 0.104

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.373 - 0.104 < p < 0.373 + 0.104

0.269 < p < 0.477

Upper bound is 0.477

The 95% confidence interval for the population proportion p is :