To determine how many samples are needed to ensure that the sample mean is between 74 and 76 with a certain probability, we need to know the population standard deviation and the desired level of confidence.
If the population standard deviation is known, we can use the formula for the margin of error:
Margin of error = z * (sigma / sqrt(n))
where z is the z-score associated with the desired level of confidence, sigma is the population standard deviation, and n is the sample size.
In this case, we don't know the population standard deviation, so we can estimate it using the sample standard deviation.
We can use the t-distribution instead of the z-distribution to account for the uncertainty in the estimate of the population standard deviation. The formula for the margin of error with the t-distribution is:
Margin of error = t * (s / sqrt(n))
where t is the t-score associated with the desired level of confidence and n is the sample size.
We want the sample mean to be between 74 and 76, so the margin of error is:
Margin of error = (76 - 74) / 2 = 1
We want to find the sample size needed to ensure that the margin of error is no more than 1 with a certain level of confidence. Let's assume a 95% confidence level, which corresponds to a t-score of 2.064 for a two-tailed test with n-1 degrees of freedom.
So we have:
1 = 2.064 * (s / sqrt(n))
Solving for n, we get:
n = (2.064 * s / 1)^2
where s is the sample standard deviation.
The sample size needed depends on the sample standard deviation. If we assume a sample standard deviation of 5 (for example), we get:
n = (2.064 * 5 / 1)^2 = 86.67
We need at least 87 samples to ensure that the sample mean is between 74 and 76 with a 95% confidence level.