Final answer:
To find the 95% confidence interval for the average weight of a nickel to within 10 milligrams, 3460 nickels need to be weighed.
Step-by-step explanation:
The subject of the question is determining the sample size needed for estimating the average weight of nickels with a predetermined precision, using confident intervals. When a coin counter manufacturer wishes to find the 95% confidence interval for the average weight of a nickel to be accurate within 10 milligrams, it implies that we want the margin of error of our estimate to be no more than 10 milligrams. To calculate the required sample size, we need to use the sample size formula for the mean, which incorporates standard deviation, margin of error, and the z-score corresponding to the desired confidence level.
To find the sample size n, we use the formula n = (z * σ / E)^2, where z is the z-score for a 95% confidence interval (which is 1.96 for a two-tailed test), σ is the standard deviation (which is 300 milligrams), and E is the desired margin of error (10 milligrams). Plugging these values into the formula gives n = (1.96 * 300 / 10)^2 = 3459.36. Since we cannot have a fraction of a sample, we round up to the next whole number, which gives us a required sample size of 3460 nickels.