1. A manufacturing firm produces a part for which the standard deviation in the diameter must not exceed 0.05 cm. to comply with the manufacturing specifications. To test that the part is in compliance, the firm randomly samples 25 parts and measures the diameter of each piece. The sampled results are to be used to test whether the part complies with the specifications. The correct null and alternative hypotheses for this test would be Ho: o20.052 vs. Ha : o<0.05² а. 2 <0.05² vs. Ha : oʻ>0.05? b. Ho: 0 20.05² vs. Ha : 0<0.05² с. d. Ho: o= 0.05² vs. Ha : o+0.05²

1. A manufacturing firm produces a part for which the standard deviation in the diameter must not exceed 0.05 cm. to comply with the manufacturing specifications. To test that the part is in compliance, the firm randomly samples 25 parts and measures the diameter of each piece. The sampled results are to be used to test whether the part complies with the specifications. The correct null and alternative hypotheses for this test would be Ho: o20.052 vs. Ha : o<0.05² а. 2 <0.05² vs. Ha : oʻ>0.05? b. Ho: 0 20.05² vs. Ha : 0<0.05² с. d. Ho: o= 0.05² vs. Ha : o+0.05²

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Description: 1. A manufacturing firm produces a part for which the standard deviation in the diameter must notexceed 0.05 cm. to comply with the manufacturing specifications. To test that the part is in compliance,the firm randomly samples 25 parts and measures

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

1. A manufacturing firm produces a part for which the standard deviation in the diameter must not
exceed 0.05 cm. to comply with the manufacturing specifications. To test that the part is in compliance,
the firm randomly samples 25 parts and measures the diameter of each piece. The sampled results are to
be used to test whether the part complies with the specifications. The correct null and alternative
hypotheses for this test would be
2
Ho: o 20.052
2
vs. Ha : o <0.05²
а.
Ho: o<0.05²
2
vs. Ha : o> 0.05²
b.
Ho: o2 0.052
vs. Ha : 0<0.05²
с.
2
vs. Ha : o 0.052
d.
Ho: o = 0.05²

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