Question

Suppose that we check for clarity in 75 locations in Lake Tahoe and discover that the average depth of clarity of the lake is 116.5 feet with a standard deviation of 2.5 feet. What can we conclude about the average clarity of the lake with a 90% confidence level? Answer 1&2

1) A) E =1.21 B) E = 0.47 C) E =1.56 D) E = 0.52

2) A) 116.03 < < μ 116.97 B) 114.26 < < μ 115.69 C) 115.29 < μ < 116.28 D) 15.63 < < μ 116.58

Answer 1

Solution :

Given that,

Z_/2 = 1.645

Margin of error = E = Z_/2* ( /n)

= 1.645 * (2.5 / 75)

E = 0.47

At 90% confidence interval estimate of the population mean is,

- E < < + E

116.5 - 0.47 < < 116.5 + 0.47

116.03 < < 116.97

(116.03 , 116.97)