Question

A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 71 specimens and counts the number of seeds in each. Use her sample results (mean = 63.3, standard deviation = 18.1) to find the 99% confidence interval for the number of seeds for the species. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places.

99% C.I. =

Answer 1

solution :

Given that,

Point estimate = sample mean = = 63.3

sample standard deviation = s = 18.1

sample size = n = 71

Degrees of freedom = df = n - 1 = 71 - 1 = 70

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = 2.648

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

= 2.648 * (18.1 / 71)

Margin of error = E = 5.688

The 99% confidence interval estimate of the population mean is,

  ± E

63.3  ± 5.688

57.612 , 68.988