Question

A statistician selected a sample of 16 accounts receivable and determined the mean of the sample to be $5,000 with a 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 he had used. Based on the above information, determine the confidence level that was used. Assume the population has a normal distribution.

Answer 1

Answer:

The confidence level that was used is 0.25% .

Step-by-step explanation:

We are given that mean of the selected sample of 16 accounts is $5,000 and a standard deviation of $400.

It has also been reported that the sample information indicated the mean of the population ranges from $4,739.80 to $5,260.20. This represents the Confidence Interval for population mean.

But we have to find that at what confidence level this information about range of population men has been stated.

Since we know that Confidence Interval for population mean is given by :

   C.I. for population mean = Sample mean(xbar) $\pm$ z value * $\frac{Standard deviation}{\sqrt{n} }$

i.e.,if we have 95% C.I. = xbar $\pm$ 1.96 * $\frac{\sigma}{\sqrt{n} }$ .

So, our Confidence Interval for population is written as :

 [$4,739.80 , $5,260.20] = $5000 $\pm$ z value * $\frac{400}{\sqrt{16} }$

 $5000 - z value * 100 = $4739.80    { Solving these we get Z value = 2.602}

 $5000 + z value * 100 = $5260.20

In z table we find that at value of 2.60 the probability is 0.99534 so subtracting this from 1 we get confidence level for one tail i.e.0.5%(approx).

Therefore, for two tail Confidence level will be 0.5%/2 = 0.25% .