A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 65 specimens and counts the number of seeds in each. Use her sample results (mean = 77.3, standard deviation = 5.5) 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.

Confidence interval = [76.4, 78.2]    to one decimal place

Step-by-step explanation:

Given:

N = size of sample = 65

m = sample mean = 77.3

s = sample standard deviation = 5.5

alpha = significance level = (100%-80%) = 20%

percentile = 100(1-alpha/2) 90%  ( assume two-sided interval)

Solution:

Confidence limits

= m +/- t(1-alpha/2, N-1) * s / sqrt(N)    t=t-distribution

= 77.3 +/- t(0.9,64) * 5.5 / sqrt(65)

= 77.3 +/- 1.29492 * 5.5 / sqrt(65)            [t-table from R]

= [76.42, 78.18]

add an image

step-by-step explanation:

14

step-by-step explanation:

14 x 3 = 42