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
[tex]t*( \frac{s}{ \sqrt{n} )} [/tex]
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.
Answer: The sample size is 81.
