If the margin of error is currently 3 with a sample size of n, how many samples do you need to collect to achieve a margin of error of 0.3?
Question Analysis
This question is about statistics, specifically concerning the concept of margin of error, which is a measure of the accuracy of a sample estimate. The margin of error is heavily dependent on the sample size. In this scenario, you are given a current margin of error of 3 with an unspecified sample size ( n ), and you need to determine the new sample size required to achieve a smaller margin of error of 0.3. This involves understanding the relationship between the margin of error and sample size, which is inversely proportional, and can be calculated using the formula for margin of error in the context of a normal distribution or a proportion.
Answer
To solve this problem, we can use the formula for the margin of error (ME) in a normal distribution:
[ \text{ME} = \frac{z \cdot \sigma}{\sqrt{n}} ]
where:
- ( z ) is the z-score (which remains constant if the confidence level is the same),
- ( \sigma ) is the standard deviation,
- ( n ) is the sample size.
Given that the initial margin of error is 3, we can express it as:
[ 3 = \frac{z \cdot \sigma}{\sqrt{n}} ]
We need to find the new sample size ( n' ) to achieve a margin of error of 0.3:
[ 0.3 = \frac{z \cdot \sigma}{\sqrt{n'}} ]
To find the new sample size ( n' ), set up the equation based on the proportional relationship of the margin of error and the inverse square root of the sample size:
[ \frac{3}{0.3} = \sqrt{\frac{n'}{n}} ]
[ 10 = \sqrt{\frac{n'}{n}} ]
Now, square both sides to solve for ( n'/n ):
[ 100 = \frac{n'}{n} ]
[ n' = 100n ]
Thus, you need to collect 100 times the current sample size ( n ) to achieve a margin of error of 0.3.