From time to time a manufacturer of pre-packed furniture needs to check the mean distance between pairs of holes drilled by a machine in pieces of chipboard to ensure that no change has occurred. It is known from experience that the standard deviation of the distance is 0.43mm. The firm intends to take a random sample of size n, and to calculate a 99% confidence interval for the mean of the population. The width of this interval must be no more than 0.60mm. Calculate the minimum value of n.
To calculate the minimum sample size (n) needed to construct a confidence interval with a given width, we can use the formula:
n = (Z * σ / E)^2
Where:
n = sample size
Z = Z-value corresponding to the desired confidence level (in this case, 99% confidence level)
σ = standard deviation
E = maximum allowable error (half the desired confidence interval width)
In this case, the confidence level is 99%, and the maximum allowable error (E) is 0.60mm / 2 = 0.30mm.
First, we need to find the Z-value corresponding to the 99% confidence level. The Z-value can be obtained from a standard normal distribution table or calculated using statistical software. For a 99% confidence level, the Z-value is approximately 2.576.
Now, we can calculate the minimum sample size (n):
n = (2.576 * 0.43 / 0.30)^2 ≈ 16.14
Since the sample size (n) must be a whole number, we round it up to the nearest integer: n = 17.
Therefore, the minimum sample size required to calculate a 99% confidence interval with a maximum width of 0.60mm is 17.