Question

From a normally-distributed population of scores with a mean of μ, 9 scores are sampled at random. The mean and standard deviation for this sample of 9 scores are found to be 12 and 4, respectively. μ is unlikely ( = :05) to be less than _______ or greater than______.

(Hint; t-dist test)

Answer 1

Solution:

Confidence interval for population mean() using t distribution  

Given that,

    n = 9 ....... Sample size

= 12 ....... Sample mean

s = 4 ........Sample standard deviation

Note that, Population standard deviation() is unknown..So we use t distribution. Given,

= 0.05

  /2 = 0.05 2 = 0.025

Also, n = 9 .....given

d.f= n-1 = 8

     =    = = 2.306

( use t table or t calculator to find this value..)

Now , confidence interval for mean() is given by:

  

12 - 2.306*(4/ 9)      12 +2.306*(4/ 9)

12-3.0747 < < 12+3.0747

8.9253 < < 15.0747

Answer: μ is unlikely ( = :05) to be less than 15.0747 or greater than 8.9253.