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
Why did the pre-packed furniture manufacturer hire a mathematician? Because they needed someone to drill down the mean distances and boardly analyze the situation!
Now, let's get serious and calculate the minimum value of n. In order to estimate the minimum sample size, we need to use the formula for the margin of error:
Margin of Error = (Z-value) * (Standard Deviation / sqrt(n))
Since the width of the confidence interval must be no more than 0.60mm, we can substitute the margin of error with 0.60mm:
0.60mm = (Z-value) * (0.43mm / sqrt(n))
To find the Z-value for a 99% confidence interval, we can use a z-table or the inverse normal function in a calculator (which is a lot more practical):
InvNorm(0.995) ≈ 2.576
Now, we can rearrange the formula and solve for n:
sqrt(n) = (Z-value * Standard Deviation) / Margin of Error
n = [(Z-value * Standard Deviation) / Margin of Error]^2
Plugging in the values:
n = [(2.576 * 0.43mm) / 0.60mm]^2
n ≈ 11.07
You can't have a fraction of a sample, so you will need a minimum sample size of 12 to ensure that the width of the confidence interval is no more than 0.60mm. Remember, it's always good to have a few extra holes... I mean, samples, just to be safe!
To calculate the minimum value of n, we need to use the formula for the width of a confidence interval:
Width = 2 * (Z * (σ/√n))
Where:
- Width is the maximum allowable width of the confidence interval (0.60mm)
- Z is the Z-value corresponding to the desired confidence level (99%)
- σ is the standard deviation of the distance between pairs of holes (0.43mm)
- n is the sample size we want to calculate
We can rearrange the formula to solve for n:
n = (2 * (Z * σ) / Width)^2
First, we need to find the Z-value corresponding to a 99% confidence level using a Z-table or Z-calculator. The Z-value for a 99% confidence level is approximately 2.576.
Now, we can substitute the values into the formula:
n = (2 * (2.576 * 0.43) / 0.60)^2
n ≈ (2.21168 / 0.60)^2
n ≈ 3.6856^2
n ≈ 13.62
So, the minimum value of n is approximately 13.62. Since we can't have a fractional sample size, we would round up to the nearest whole number. Therefore, the minimum value of n would be 14.
To calculate the minimum value of n, we need to use the formula for the margin of error in a confidence interval:
Margin of Error = Z * (Standard Deviation / √n)
Where:
Z is the Z-score corresponding to the desired level of confidence (99% in this case).
Standard Deviation is the known standard deviation of the population (0.43mm).
n is the sample size.
We also know that the width of the confidence interval must be no more than 0.60mm. The width of a confidence interval is given by the formula:
Width = 2 * Margin of Error
Therefore, we can rewrite the margin of error formula as:
Margin of Error = Width / 2
Substituting the given values, we have:
Width = 0.60mm
Standard Deviation = 0.43mm
With these values, we can now calculate n.
First, let's calculate the margin of error:
Margin of Error = Width / 2 = 0.60mm / 2 = 0.30mm
Next, we need to find the Z-score corresponding to a 99% confidence level. The Z-score can be found using a standard normal distribution table or a Z-score calculator. For a 99% confidence level, the Z-score is approximately 2.58.
Now, we can plug in the values into the margin of error formula:
0.30mm = 2.58 * (0.43mm / √n)
To solve for n, we can isolate it on one side of the equation:
√n = (2.58 * 0.43mm) / 0.30mm
√n = 3.6806
Square both sides of the equation:
n = (3.6806)^2
n ≈ 13.55 (rounded up to the nearest whole number)
Therefore, the minimum value of n is 14 (to ensure a sample size greater than or equal to 13.55) in order to meet the criteria of a 99% confidence interval with a maximum width of 0.60mm.