The owner of a trout farm routinely measures the lengths of his trouts. He randomly took...

The owner of a trout farm routinely measures the lengths of his trouts. He randomly took 31 trouts last Friday from his pond and measured their lengths. The mean length was 26.1 centimeters with a standard deviation of 1.8. Estimate the true mean length of all trouts in his farm using a 90% confidence interval.

(a) The limits of the confidence interval are calculated by x¯±tsn‾√x¯±tsn , where

x¯=

t= (Round to 3 decimal places.) Please explain how to find this!

(b) The 90% confidence interval for the true mean length is: ( , ) (Round to 3 decimal places.)

Solutions

Expert Solution

Solution :

Given that,

= 26.1

s = 1.8

n = 25

Degrees of freedom = df = n - 1 = 31 - 1 = 30

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t_/2,df = t_0.05,30 = 1.697

Margin of error = E = t_/2,df * (s / $\sqrt$ n)

= 1.697 * ( 1.8 / $\sqrt$ 31)

= 0.549

Margin of error = 0.549

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

- E < < + E

26.1 - 0.549 < < 26.1 + 0.549

25.551 < < 26.649

(25.551, 26.649 )

orchestra answered 1 year ago

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