Question

3. The Hammermill Company produces paper for laser printers. Standard paper width is 216mm, or 8,5 inches. Suppose that the actual width is a random variable that is normally distributed with a known standard deviation of .0230 based on the manufacturing technology currently in use. Variation arises during manufacturing because of slight differences in the paper stock, vibration in the rollers and cutting tools, and wear and tear on the equipment. The cutters can be adjusted if the paper width drifts from the correct mean. Suppose that a quality control inspector chooses 50 sheets at random and measures them with a precise instrument, obtaining a mean width of 216.0070 mm. Using a 5 percent level of significance (a = .05), does this sample show that the product mean exceeds the specification?

Answer 1

Solution:

Here, we have to use one sample z test for population mean.

H₀: µ = 216 versus H_a: µ > 216

This is an upper tailed test.

WE are given

Level of significance = α = 0.05

Sample mean = Xbar = 216.0070

Sample size = n = 50

Population standard deviation = σ = 0.0230

Critical Z value = 1.6449

(by using z-table)

Test statistic formula is given as below:

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (216.0070 – 216)/[0.0230/sqrt(50)]

Z = 2.1521

P-value = 0.0157

(by using z-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that product mean exceeds the specification.