PLLZZZZ HELP I WILL GIVE 50 POINTS
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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><em> z value * </em>
<em />
i.e.,<em>if we have 95% C.I. = xbar </em><em> 1.96 * </em>
<em> .</em>
So, our Confidence Interval for population is written as :
[$4,739.80 , $5,260.20] = $5000 z value *
$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% .