Question

11. A random sample of 50 college student students shows that the score of a College Statistics is normally distributed with its mean, 83 and standard deviation, 7.5. Find 95 % confidence interval estimate for the true mean.

Answer 1

Solution :

Given that,

Point estimate = sample mean =     = 83

Population standard deviation =    = 7.5

Sample size n =50

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E =   Z/2    * ( /n)
= 1.96 * ( 7.5/ 50)

= 2.08
At 95% confidence interval estimate of the population mean
is,

- E < < + E

83 - 2.08 <   < 83 + 2.08

80.92 <   < 85.08