Question

A sample of 9 adult elephants had an average weight of 11,773 pounds. The standard deviation for the sample was 24 pounds. Find the 99% confidence interval of the population mean for the weights of adult elephants. Assume the variable is normally distributed. Round intermediate answers to at least three decimal places. Round your final answers to the nearest whole number.

Answer 1

99% C.I of the population mean is ( 11746, 11800 )

Explanation:

Given the data in the question;

Sample size n = 9

mean x" = 11773

standard deviation s = 24

for 99% confidence interval

∝ = 1 - 99% = 1 - 0.99 = 0.01

∝/2 = 0.01/2 = 0.005

degree of freedom df = n - 1 = 9 - 1 = 8

now, Critical t-value :
=
= 3.355

Now, at 99% Confidence interval of the mean population is;

x" ± [
× s/√n ]

we substitute

11773 ± [ 3.355 × 24/√9 ]

11773 ± [ 3.355 × 8 ]

11773 ± 26.84

so

⇒ {11773 - 26.84}, {11773 + 26.84}

⇒ ( 11746, 11800 )

Therefore, 99% C.I of the population mean is ( 11746, 11800 )