Confidence Level Example

A statistician selected a sample of 16 accounts receivable and determined the mean of the sample to be $5,000 with a sample standard deviation of $400. He reported that the sample information indicated the mean of the population ranges from $4,739.80 to $5,260.20. He neglected to report what confidence level (1-α) he had used. Based on the above information, determine the confidence level that was used.

Answer:

The statistician is reporting a confidence interval of 5000 ± 260.20. He only mentions the sample standard deviation (not the population standard deviation), so he must be using the t-distribution and the formula:

 ± t_{n-1, α/2}(s/√n).

So we have:
260.2 = t(s/√n)
260.2 = t(400/√16)
260.2 = 100t
t = 2.602

We look to the t distribution table and find that t_{15, α/2} = 2.602 is true for α/2 = 0.01. So α = 0.02 and the confidence level is 1-0.02 = 0.98 = 98%.

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