Question

Two hundred fish caught in Cayuga Lake had a mean length of 14.8 inches. The population standard deviation is 3.7 inches. (Give your answer correct to two decimal places.)

(a) Find the 90% confidence interval for the population mean length.

Lower Limit

Upper Limit

(b) Find the 98% confidence interval for the population mean length.

Lower Limit

Upper Limit

Answer 1

Solution :

Given that,

= 14.8

= 3.7

n = 200

(A)

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* ( /n)

= 1.645* (3.7 / 200)

E= 0.43

At 90% confidence interval for the population mean length.,

- E < < + E

14.8-0.43 < < 14.8+0.43

14.37< < 15.23

(14.37,15.23 )

Lower Limit 14.37

Upper Limit 15.23

(B)

Solution :

Given that,

= 14.8

= 3.7

n = 200

(A)

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z_/2 = Z_0.01 = 2.326

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

= 2.326* (3.7 / 200)

E= 0.61

At 98% confidence interval for the population mean length.

- E < < + E

14.8-0.61< < 14.8+0.61

14.19< < 15.41

(14.19,15.23 )

Lower Limit 14.19

Upper Limit 15.41