Find a number 't' such that the probability that the sample mean will differ from the population mean by at least 't' is not greater than 0.05 if the sample size is 100 and sigma is 3
To find the number 't' that satisfies this condition, we need to calculate the margin of error. The margin of error is the maximum amount by which the sample mean is allowed to differ from the population mean.
The formula for the margin of error is:
Margin of error = z * (sigma / sqrt(n))
In this formula:
- 'z' represents the z-score corresponding to the desired level of confidence. For a 95% confidence level (which corresponds to a probability greater than 0.05), the z-score is approximately 1.96.
- 'sigma' is the population standard deviation.
- 'n' is the sample size.
Substituting the given values into the formula:
Margin of error = 1.96 * (3 / sqrt(100))
= 1.96 * (3 / 10)
= 1.96 * 0.3
= 0.588
The number 't' we are looking for is the margin of error. So, 't' is approximately 0.588. This means that the probability that the sample mean will differ from the population mean by at least 0.588 (or greater) is not greater than 0.05 for a sample size of 100 and a population standard deviation of 3.