Final answer:
To calculate the sample size (n) required for a 95% confidence interval with a length of 60, we need to know the population standard deviation (σ). With σ and the z-score for 95% confidence level, we can use the formula EBM = (z * σ) / √n to find the sample size that satisfies the desired margin of error of 30 units (half the interval width).
Step-by-step explanation:
The question pertains to the calculation of a sample size (n) needed to estimate the population mean gain with a specified margin of error in a confidence interval (CI). A confidence interval is a range within which we expect the true population parameter to fall with a certain level of confidence. In this scenario, the student desires a 95% CI with a total width, or length, of 60 units for the true mean gain. This implies that the desired margin of error (the maximum difference between the population mean and the sample mean) is 30 units (half of the total width).
To construct a 95% confidence interval, we use the formula (x - EBM, x + EBM), where x is the sample mean and EBM is the error bound for the population mean, which depends on the standard deviation of the population (σ), the desired confidence level (which affects the z-score), and the sample size (n).
Since the standard deviation (σ) is known and the desired EBM is 30, we can use the equation EBM = (z * σ) / √n to solve for n. However, the value of the standard deviation (σ) is not provided in the question. Given the value of σ and the z-score for the 95% confidence level (typically 1.96), the sample size (n) can be calculated accordingly to ensure the length of the interval is 60 units.