A candy machine distributor collects the coins from 100 machines. The standard deviation is $3.25. With probability 98%, what is the maximum error about the true mean that will result?
98% = mean ± 2.325 SEm
SEm = SD/√n
=2.325(3.25/ square root of 100)
=75.5625
To calculate the maximum error about the true mean, we need to use the concept of confidence intervals. A confidence interval indicates the range within which the true population parameter is likely to fall with a certain level of confidence.
In this case, we want to find the maximum error about the true mean with a 98% confidence level.
The formula to calculate the maximum error about the mean (also known as the margin of error) is:
Margin of Error = Z * (Standard Deviation / √n)
Where:
Z is the z-score corresponding to the desired confidence level
Standard Deviation is the standard deviation of the population
n is the sample size
In our case, we know that the standard deviation is $3.25, and we have collected data from 100 machines (n = 100). The z-score corresponding to a 98% confidence level is 2.33.
Now let's plug the values into the formula:
Margin of Error = 2.33 * ($3.25 / √100)
First, we calculate the square root of the sample size:
√100 = 10
Now we can calculate the margin of error:
Margin of Error = 2.33 * ($3.25 / 10)
After performing the calculations, we find that the maximum error about the true mean is approximately $0.75.
Therefore, with a 98% confidence level, the maximum error about the true mean in collecting coins from 100 machines would be $0.75.