Question

A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 60 specimens and counts the number of seeds in each. Use her sample results (mean = 17.1, standard deviation = 16.4) to find the 95% confidence interval for the number of seeds for the species. Enter your answer as an open-interval (i.e., parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 17.1

sample standard deviation = s = 16.4

sample size = n = 60

Degrees of freedom = df = n - 1 = 60 - 1 = 59

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t _{0.025, 59} = 2.001

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

= 2.001 * ( 16.4/ 60)

Margin of error = E = 4.2

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

  ± E

= 17.1 ± 4.2

=( 12.9, 21.3 )