26) Use the given info. To find the minimum sample size required to estimate an unknown population mean “M”.?
26) Use the given info. To find the minimum sample size required to estimate an unknown population mean “M”.
How many weeks of data must be randomly sampled to estimate the mean weekly sales of a new line of athletic footwear? We want 98% confidence that the sample mean is within $400 if the population mean, and the population standard deviation is known to be $1400.
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26) ANSWER: Sample Size = 67 for 98% level of confidence
Why???
SMALL SAMPLE, LEVEL OF CONFIDENCE, NORMAL POPULATION DISTRIBUTION
Margin of Error (half of confidence interval) = 400
The margin of error is defined as the "radius" (or half the width) of a confidence interval for a particular statistic.
Level of Confidence = 98
σ: population standard deviation = 1400
('z critical value') from Look-up Table for 98% = 2.326
The Look-up in the Table for the Standard Normal Distribution utilizes the Table's cummulative 'area' feature. The Table shows positve and negative values of ('z critical') but since the Standard Normal Distribution is symmetric, the magnitude of ('z critical') is important.
For a Level of Confidence = 98% the corresponding LEFT 'area' = 0.49. And due to Table's symmetric nature, the corresponding RIGHT 'area' = 0.49 The ('z critical') value Look-up is 2.326 which means the MIDDLE 'area' = SUM[LEFT 'area' + RIGHT 'area'] for a Level of Confidence = 98.
significant digits = 3
Margin of Error = ('z critical value') * σ/SQRT(n)
n = Sample Size
Algebraic solution for n:
n = [('z critical value') * σ/Margin of Error]²
= [ (2.326 * 1400)/400 ]²
Sample Size = 67 for 98% level of confidence