Question

A random sample of 64 drivers used on average 747 gallons of gasoline per year. The standard deviation of the population is 34 gallons.

Find the 99% confidence interval of the mean for all drivers. Round intermediate answers to at least three decimal places. Round your final answers to the nearest whole number.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 747

Population standard deviation =    = 34

Sample size = n =64

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576   ( Using z table )

Margin of error = E = Z/2* ( /n)

= 2.576* (34 / 64)

E= 10.9480

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

- E < < + E

747- 10.9480< <747+ 10.9480

736.0520< < 757.9480

(736., 758)