Statistics: Confidence intervas and sample size?

A researcher wants to estimate a population mean (of serum cholesterol level of a certain population of middle-aged men) to within +/- 6mg/dLi or less with a 95% confidence. The standard error of the mean should be 3mg/dLi or less. Also, the resaecher believes the standard deviation of serum cholesterol in the population is probably about 40mg/dLi. How large a sample does the researcher need to take?





And explanation would be pretty helpful along with an answer.|||ANSWER: Sample Size = 171 for 95% level of confidence





Why???





SMALL SAMPLE, LEVEL OF CONFIDENCE, NORMAL POPULATION DISTRIBUTION








Margin of Error (half of confidence interval) = 6


The margin of error is defined as the "radius" (or half the width) of a confidence interval for a particular statistic.


Level of Confidence = 95


蟽: population standard deviation = 40


('z critical value') from Look-up Table for 95% = 1.96


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 = 95% the corresponding LEFT 'area' = 0.475. And due to Table's symmetric nature, the corresponding RIGHT 'area' = 0.475 The ('z critical') value Look-up is 1.96 which means the MIDDLE 'area' = SUM[LEFT 'area' + RIGHT 'area'] for a Level of Confidence = 95.

















significant digits = 3





Margin of Error = ('z critical value') * 蟽/SQRT(n)


n = Sample Size





Algebraic solution for n:


n = [('z critical value') * 蟽/Margin of Error]虏


= [ (1.96 * 40)/6 ]虏





Sample Size = 171 for 95% level of confidence