Question

Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean: 18, 12, 20, 17, 14, 15, 13, 11, 21, 17. Assume that the population germination time is normally distributed. Find the 99% confidence interval for the mean germination time.

Answer 1

From given sample ,

= X / n = 15.8

S = sqrt [ X² - n ² / n-1 ] = 3.3599

t critical value at 0.01 significance level with 9 df = 3.250

99% confidence interval for is

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

15.8 - 3.25 * 3.3599 / sqrt(10) < < 15.8 + 3.25 * 3.3599 / sqrt(10)

12.35 < < 19.25

99% CI is ( 12.35 , 19.25 )