A sample of 9 adult elephants has an average weight of 12,500 pounds. The standard deviation...

A sample of 9 adult elephants has an average weight of 12,500 pounds. The standard deviation for the sample was 24 pounds. Find the 99% confidence interval of the population mean for the weight of adult elephants. Assume the variable is normally distributed. Round answer to the nearest whole number.

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Expert Solution

solution

Given that,

= 12500

s =24

n = 9

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

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2,df= t0.005, 8= 3.355 ( using student t table)

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

=3.355 * (24 / $\sqrt$ 9) = 26.8400

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

- E < < + E

12500 -26.8400 < < 12500+ 26.8400

12473 < < 12527

(12473 , 12527 )

orchestra answered 1 year ago

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