Question

Two hundred fish caught in Cayuga Lake had a mean length of 13.8 inches. The population standard deviation is 3.8 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,

= 13.8

= 3.8

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 * (3.8 / 200 ) = 0.442

At 95% confidence interval estimate of the population mean is,

- E < < + E

13.8 - 0.442< < 13.8  + 0.442

13.36 < <14.24

(13.36 , 14.24)

Lower limit - 13.36

Upper limit - 14.24

b) 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)

= 1.326 * (3.8 / 200) = 0.625

At 98% confidence interval estimate of the population mean is,

- E < < + E

13.8 - 0.625 < < 13.8 + 0.625

13.18 < < 14.42

(13.18 , 14.42)

Lower limit - 13.18

Upper limit - 14.42