Lumber Retail
STATISTICS Question about HOW LARGE a sample is needed?????
A retail lumberyard routinely inspects incoming shipments of lumber from suppliers. For select grade 8-foot 2 -by-4 pine shipments, the lumberyard supervisor chooses one gross (144 boards) randomly from a shipment of several tens of thousands of boards. In the sample 18 boards are not salable as select grade.
a) The lumberyard must decide how many boards per shipment will be inspected. What sample size is needed to obtain a 95% confidence interval for the proportion of unsalable boards with a width of .04? Assume that somewhere between 10% and 20% of a shipment is unsalable.
b) In this situatio, would it be sensible to calculate a sample size based on the worst-case assumption that 50% of the shipment is unsalable?
Confidence intervals are used to find a region in which we are 100 * ( 1 – α )% confident the true value of the parameter is in the interval.
For large sample confidence intervals about the population proportion you have:
pHat ± z * sqrt(phat * (1- phat) / n)
where phat is the sample proportion
z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α
n is the sample size
To find the sample size needed for a confidence interval of a given size we need only to concern ourselves with the error term and the width of the interval.
We know that the interval is centered at phat so we need to find the value of n such that
z * sqrt(phat * (1-phat) / n) = width.
The z-score for a 0.95 confidence interval is the value of z such that 0.025 is in each tail of the distribution.
z= 1.959964
The equation we need to solve is: z * sqrt(phat * (1-phat) / n) = width
n = phat * (1 – phat) * (width / z) ^ -2.
If we don’t know anything about phat and are still asked to find the sample size we let phat = 0.5. This maximizes the value of the error term and if n is sufficient for phat = 0.5, the n will be sufficient for all other values of phat.
to find the largest sample needed with the assumption that the proportion is between 10% and 20% I will use the 20%.
n = 0.2 * ( 1 – 0.2 ) * ( 0.04 / 1.959964 ) ^ -2
n = 384.1459
n must be integer valued. Always take the ceiling of n so that the size of the interval will be correct.
n = 385
if we use phat = 0.5 we have the following:
n = 0.5 * ( 1 – 0.5 ) * ( 0.04 / 1.959964 ) ^ -2
n = 600.228
n must be integer valued. Always take the ceiling of n so that the size of the interval will be correct.
n = 601
Lakeside Lumber – Ashby, MN
