Question

A sample of the reading scores of 35 fifth-graders has a mean of 82. The standard deviation of the population is 15. Find the 99% confidence interval of the mean reading scores of all fifth-graders.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 82

Population standard deviation =    = 15
Sample size = n =35

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 * (15 / 35)

= 6.5314

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

- E < < + E

82- 6.5314< < 82+ 6.5314

75.4686< < 88.5314

(75.4686, 88.5314 )