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insens350 [35]
3 years ago
9

A statistician selected a sample of 16 accounts receivable and determined the mean of the sample to be $5,000 with a standard de

viation 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.
Mathematics
1 answer:
Anon25 [30]3 years ago
4 0

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.

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

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

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

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

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% .

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