Number of nickels needed when only standard deviation is mentioned

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SUMMARY

The discussion focuses on determining the required sample size of nickels to achieve a 98% confidence interval for the average weight, given a standard deviation of 150 milligrams and a desired margin of error of 15 milligrams. The appropriate sample size formula for this scenario is derived from the formula for the margin of error in estimating a population mean, which is calculated as \( n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \). Here, \( Z \) is the Z-score corresponding to the 98% confidence level, \( \sigma \) is the standard deviation, and \( E \) is the margin of error.

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  • Understanding of confidence intervals and their significance
  • Familiarity with standard deviation and its role in statistics
  • Knowledge of sample size determination formulas
  • Basic proficiency in statistical concepts such as Z-scores
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#of nickels needed when only standard deviation is mentioned

A previous study of nickels showed that the the standard deviation of the weight of nickels is 150 milligrams. A coin counter manufacturer wishes to find the 98% confidence interval for the average weight of a nickel. How many nickels does he need to weigh to be accurate within 15 milligrams?

Please help I don't know where to start this question except that I know that it has a standard deviation of 150.
 
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Think this way: "... accurate to within 15 milligrams?" can be interpreted as saying you want to construct a 98% confidence interval for the mean weight, and you want to have a large enough sample size for the margin of error to be 15 milligrams. How large a sample is needed. What sample size formula is appropriate?
 

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