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a survey was conducted of the age (in years) of 30 randomly selected customers. The mean was 32.52 years and the standard deviation was 11.81 years. a) How large would the sample size have to be to cut the margin of error in half? b) The new sample size should be?
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so margin of error now is z*32.52/sqrt(30). so n has to equal half this value. z*32.52/2sqrt(30) = z*32.52/sqrt(n) if you solve for n, it'll be 120. (check my calculation just in case.)
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