Question

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.)

s=

n=

(b) The 90% confidence interval for the true mean length is:

(  ,  ) (Round to 3 decimal places.)

Answer 1

a)

= 26.1

t = 1.697 (With 30 df and at 0.10 significance level)

S = 1.8

n = 31

b)

90% confidence interval for is

- t * S / sqrt(n) < < + t * S / sqrt(n)

26.1 - 1.697 * 1.8 / sqrt(31) < < 26.1 + 1.697 * 1.8 / sqrt(31)

25.551 < < 26.649

90% CI is ( 25.551 , 26.649)