Question

A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 64 specimens and counts the number of seeds in each. Use her sample results (mean = 73.2, standard deviation = 13.2) to find the 80% 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).

80% C.I. =

Answer should be obtained without any preliminary rounding.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 73.2

sample standard deviation = s = 13.2

sample size = n = 64

Degrees of freedom = df = n - 1 = 64 - 1 = 63

At 80% confidence level

= 1 - 80%

=1 - 0.80 =0.20

/2 = 0.10

t/2,df = t0.10,63 = 1.295

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

= 1.295 * (13.2 / 64)

Margin of error = E = 2.1

The 80% confidence interval estimate of the population mean is,

  ± E  

= 73.2  ± 2.1

= ( 71.1, 75.3 )