<span>n = 5 The formula for the confidence interval (CI) is CI = m Â± z*d/sqrt(n) where CI = confidence interval m = mean z = z value in standard normal table for desired confidence n = number of samples Since we want a 95% confidence interval, we need to divide that in half to get 95/2 = 47.5 Looking up 0.475 in a standard normal table gives us a z value of 1.96 Since we want the margin of error to be Â± 0.0001, we want the expression Â± z*d/sqrt(n) to also be Â± 0.0001. And to simplify things, we can omit the Â± and use the formula 0.0001 = z*d/sqrt(n) Substitute the value z that we looked up, and get 0.0001 = 1.96*d/sqrt(n) Substitute the standard deviation that we were given and 0.0001 = 1.96*0.001/sqrt(n) 0.0001 = 0.00196/sqrt(n) Solve for n 0.0001*sqrt(n) = 0.00196 sqrt(n) = 19.6 n = 4.427188724 Since you can't have a fractional value for n, then n should be at least 5 for a 95% confidence interval that the measured mean is within 0.0001 grams of the correct mass.</span>

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3 years ago