Question

Lazurus Steel Corporation produces iron rods that are supposed to be 36 inches long. The machine that makes these rods does not produce each rod exactly 36 inches long. The lengths of the rods are normally distributed and vary slightly. It is known that when the machine is working properly, the mean length of the rods is 36 inches. The standard deviation of the lengths of all rods produced on this machine is always equal to 0.035 inch. The quality control department at the company takes a sample of 20 such rods every week, calculates the mean length of these rods, and tests the null hypothesis, μ=36 inches, against the alternative hypothesis, μ≠36 inches. If the null hypothesis is rejected, the machine is stopped and adjusted. A recent sample of 20 rods produced a mean length of 36.015 inches.

Calculate the p-value for this test of hypothesis. Based on this p-value, will the quality control inspector decide to stop the machine and adjust it if he chooses the maximum probability of a Type I error to be 0.01?

Use the normal distribution table. Round your answer to four decimal places.

p-value =

The machine

does not need or needs

adjustment.

the tolerance is +/-2%

Answer 1

Solution:

Given: When the machine is working properly, the mean length of the rods is 36 inches.

thus

The standard deviation of the lengths of all rods produced on this machine is always equal to 0.035 inch.

that is:

the null hypothesis, H0: μ=36 inches, against the alternative hypothesis, H1: μ≠36 inches.

If the null hypothesis is rejected, the machine is stopped and adjusted.

Sample size = n = 20

Sample mean =

We have to calculate the p-value for this test of hypothesis.

Level of significance =

Find z test statistic:

Since H1 is ≠ type , this is two tailed test, p-value is given by:

For two tailed test , p-value is:

p-value = 2* P(Z > z test statistic) if z is positive

p-value = 2* P(Z < z test statistic) if z is negative

thus

p-value = 2* P(Z > z test statistic)

p-value = 2* P(Z > 1.92)

p-value = 2* [ 1 - P(Z < 1.92) ]

Look in z table for z = 1.9 and 0.02 and find corresponding area.

P( Z < 1.92) = 0.9726

thus

p-value = 2* [ 1 - P(Z < 1.92) ]

p-value = 2* [ 1 - 0.9726 ]

p-value = 2* 0.0274

p-value = 0.0548

Since p-value = 0.0548 > 0.01 level of significance, we fail to reject H0.

Thus

The machine does not need  adjustment