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.

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) [tex]\pm[/tex] z value * [tex]\frac{Standard deviation}{\sqrt{n} }[/tex]

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

So, our Confidence Interval for population is written as :

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

$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% .

joaobezerra joaobezerra · Answer 2 · 2021-11-16T17:32:20+01:00

Using the t-distribution, it is found that a confidence level of 98% was used.

We are given the standard deviation for the sample, which is why the t-distribution is used to solve this question.

The information given is:

Sample mean of [tex]\overline{x} = 5000[/tex].
Sample standard deviation of [tex]s = 400[/tex].
Sample size of [tex]n = 16[/tex].

The margin of error is of:

[tex]M = t\frac{s}{\sqrt{n}}[/tex]

In which t is the critical value.

For this problem, the margin of error is:

[tex]M = \frac{5260.2 - 4739.8}{2} = 260.2[/tex]

Hence, the critical value is found solving the equation of the margin of error for t.

[tex]M = t\frac{s}{\sqrt{n}}[/tex]

[tex]260.2 = t\frac{400}{\sqrt{16}}[/tex]

[tex]100t = 260.2[/tex]

[tex]t = \frac{260.2}{100}[/tex]

[tex]t = 2.602[/tex]

Looking at the t-table, with 16 - 1 = 15 df, t = 2.602 is associated with a confidence level of 98%.

A similar problem is given at https://brainly.com/question/15180581