Question

a random sample of 65 mature redwood trees yields a sample mean diameter of 184.3 cm and a sample standard deviation of 19.9 cm. At the 99% confidence level, find the margin of error(also called the maximum error) of the estimate of this mean to the nearest tenth.

Answer 1

Solution :

Given that,

= 184.3

s = 19.9

n = 65

Degrees of freedom = df = n - 1 = 65 - 1 = 64

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2,df = t0.005,64 =2.654

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

= 2.797 * (19.9 / 65)

= 52.8

Margin of error = 52.8