Question

Wildlife conservationist studying grizzly bears in the United States found that the mean weight of 25 adult males was 600 pounds with a standard deviation of 90 pounds. construct and interpret a 98% confidence interval for the mean weight of all adult male grizzly bears in the United States. Assume that the weights of all adukt male grizzly bears in the United States are normally distributed. Round to 3 decimal places.

Answer 1

solution :

Given that,

Point estimate = sample mean = = 600

sample standard deviation = s = 90

sample size = n = 25

Degrees of freedom = df = n - 1 = 25 - 1 = 24

At 98% confidence level

= 1 - 98%

=1 - 0.98 = 0.02

/2 = 0.01

t/2,df = 2.492

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

= 2.492 * (90 / 25)

Margin of error = E = 44.856

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

  ± E

600  ± 44.856

(555.144 , 644.856)