In an automobile repair and service industry, the average number of defective items in all the lots of automobile parts produced by a machine last week was 8. The supervisor believes that the mean number of defective parts produced by the same machine will differ in the present week. To test whether the average differs from 8, he samples 30 lots to obtain the mean number of defective parts to be 9 with a standard deviation of 3.
In an automobile repair and service industry, the average number of defective items in all the lots of automobile parts produced by a machine last week was 8. The supervisor believes that the mean number of defective parts produced by the same machine will differ in the present week. To test whether the average differs from 8, he samples 30 lots to obtain the mean number of defective parts to be 9 with a standard deviation of 3.
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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In an automobile repair and service industry, the average number of defective items in all the lots of automobile parts produced by a machine last week was 8. The supervisor believes that the mean number of defective parts produced by the same machine will differ in the present week. To test whether the average differs from 8, he samples 30 lots to obtain the mean number of defective parts to be 9 with a standard deviation of 3.
Using the critical value approach, identify the decision at the 5% level of significance.
Using the critical value approach, identify the decision at the 5% level of significance.
Group of answer choices
The decision is not to reject H0 because the test statistic does not fall in the rejection region (-∞,-1.960] U [1.960, ∞).
The decision is to reject H0 because the test statistic falls in the rejection region (-∞,-1.645].
The decision is not to reject H0 because the test statistic does not fall in the rejection region [1.645, ∞).
The decision is not to reject H0 because the test statistic does not fall in the rejection region (-∞,-1.645] U [1.645, ∞).
The decision is to reject H0 because the t
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