Lazurus Steel Corporation produces iron rods that are supposed to be 36 inches l
ID: 3175086 • Letter: L
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, =36 inches, against the alternative hypothesis, 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 0.025?
Use the normal distribution table. Round your answer to four decimal places.
p-value =
The machine: needs or does not need, adjustment.
the tolerance is +/-5%
Explanation / Answer
Given that,
Standard deviation, =0.035
Null, H0: =36
Alternate, H1: !=36
Level of significance, = 0.025
From Standard normal table, Z /2 =2.24
Since our test is two-tailed
Reject Ho, if Zo < -2.24 OR if Zo > 2.24
Reject Ho if (x-36)/0.035/(n) < -2.24 OR if (x-36)/0.035/(n) > 2.24
Reject Ho if x < 36-0.078/(n) OR if x > 36-0.078/(n)
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Suppose the size of the sample is n = 20 then the critical region
becomes,
Reject Ho if x < 36-0.078/(20) OR if x > 36+0.078/(20)
Reject Ho if x < 35.982 OR if x > 36.018
Suppose the true mean is 36.015
Probability of Type I error,
P(Type I error) = P(Reject Ho | Ho is true )
= P(35.982 < x OR x >36.018 | 1 = 36.015)
= P(35.982-36.015/0.035/(20) < x - / /n OR x - / /n >36.018-36.015/0.035/(20)
= P(-4.217 < Z OR Z >0.383 )
= P( Z <-4.217) + P( Z > 0.383)
= 0 + 0.3509 [ Using Z Table ]
= 0.3509
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