A County Water Department tested the drinking water of randomly selected 10 homeowners for lead and copper contamination. The test results yielded sample mean and sample standard deviation for lead as 2.89 and 3.92 (in µg/L), respectively, and for copper as 0.4083 and 0.2495 (in mg/L), respectively. a) Set up 90% confidence interval for the mean lead level in drinking water in that county and interpret the result. (3 Points) b) Set up 90% confidence interval for the mean copper level in drinking water in that county and interpret the result. (3 Points) c) What are the underlying assumptions? (1 Point)

a. (0.6178, 5.1622)

b. (0.2637, 0.5592)

c.

Step-by-step explanation:

A.

For lead

n1 = 10

X1 = 2.89

S1 = 3.92

90% ci = 1-90%

= 0.10

TCritical = 1.833

Degree of freedom = 10-1 = 9

X1 +- TCritical * s/√n

= 2.88 +-1.833x3.92/√10

= 2.89+-2.2722

= 0.6178, 5.1622

Lead in water lies between this interval at 90% confidence level.

For cooper

n2 = 10

X2 = 0.4083

S2 = 0.2495

Alpha = 0.10

Df = 9

TCritical = 1.833

0.4083+-1.833*9.2495/√10

= 0.4084+-0.14462

= (0.26368, 0.55292)

Copper in water lies between this interval at 90% confidence level

C.

Underlying assumptions

The samples are randomly taken as well as independent. The sample size is not big enough. So it is approximately normally distributed.