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Question 893:



No answer provided yet.The margin of error is half the width of the confidence interval. So we need to construct a confidence interval using the t-distribution for a 99% level of confidence.

First we need to construct the standard error of the mean (SEM). The SEM is how much we'd expect a sample mean to fluctuate given the sample size and standard deviation. It is found by dividing the standard deviation by the square root of the sample size: SEM = s/SQRT(n) and equals = .70/SQRT(32) = .1237.

The margin of error is the SEM times the critical value from the t-distribution for the specified level of confidence. This can be found using a t-table or the excel function =TINV(.01,31) where the parameters in the formula are 1-confidence level and the degrees of freedom (n-1). We get the critical value of 2.744.

Therefore the margin of error is .1237*2.744 = .33955 or about 2.3% of the mean.

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