Solution Found!
Linear Rotary Bearing A linear rotary bearing is designed
Chapter 6, Problem 11RE(choose chapter or problem)
A linear rotary bearing is designed so that the distance between the retaining rings is 0.875 inch. The quality-control manager suspects that the manufacturing process needs to be recalibrated because the mean distance between the retaining rings is greater than 0.875 inch. In a random sample of 36 bearings, he finds the sample mean distance between the retaining rings is 0.876 inch with standard deviation 0.005 inch.
(a) Are the requirements for conducting a hypothesis test satisfied?
(b) State the null and alternative hypotheses.
(c) The quality-control manager decides to use an \(\alpha=0.01\) level of significance. Why do you think this level of significance was chosen?
(d) Does the evidence suggest the machine be recalibrated?
(e) What does it mean for the quality-control engineer to make a Type I error? A Type II error?
Questions & Answers
QUESTION:
A linear rotary bearing is designed so that the distance between the retaining rings is 0.875 inch. The quality-control manager suspects that the manufacturing process needs to be recalibrated because the mean distance between the retaining rings is greater than 0.875 inch. In a random sample of 36 bearings, he finds the sample mean distance between the retaining rings is 0.876 inch with standard deviation 0.005 inch.
(a) Are the requirements for conducting a hypothesis test satisfied?
(b) State the null and alternative hypotheses.
(c) The quality-control manager decides to use an \(\alpha=0.01\) level of significance. Why do you think this level of significance was chosen?
(d) Does the evidence suggest the machine be recalibrated?
(e) What does it mean for the quality-control engineer to make a Type I error? A Type II error?
ANSWER:Step 1 of 4
From the given information
\(\mu=0.875 \text { inch, } \bar{x}=0.876 \text { inch, } \mathrm{s}=0.005 \text { inch , } \mathrm{n}=36 \text { bearings. }\)