Determine the sample size required to estimate the mean score on a standardized test within 5 points of the true mean with 99% confidence. assume that s=17 based on earlier studies.

Given:
Accuracy = 5
99% confidence interval
s = 17, sample standard deviation.

Because the population standard deviation is unknown, we should use the Student's t distribution.
The accuracy at the 99% confidence level for estimating the true mean is

where
n =  the sample size.
t* is provided by the t-table.

That is,
(17t*)/√n = 5
√n = (17t*)/5 = 3.4t*
n = 11.56(t*)²

A table of t* values versus df (degrees of freedom) is as follows.
Note that df = n-1.

     n      df         t*
------ --------  -------
1001 1000  2.581
  101    100  2.626
   81      80  2.639
   61      60  2.660

We should evaluate iteratively until the guessed value, n, agrees with the computed value, N.

Try n = 1001 => df = 1000.
t* = 2.581
N = 11.56*(2.581²) = 77
No agreement.

Try n = 81 => df = 80
t* = 2.639
N = 11.56*(2.639²) = 80.5
Good agreement

We conclude that n = 81.

The sample size is 81.

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