An automatic coffee filling process is in control when the average amount of instant coffee that is poured into a jar is 8 oz and the standard deviation is 0.2 oz. A simple random sample of 100 jars is selected and the sample average is found to be X = 8.1 Oz. Assuming the standard deviation of the process is still 0.2 oz, is the process out of control (pouring too much coffee into the jars) after specifying the following: (i) State your null and alternative hypotheses. (ii) State your level of significance. (iii) Specify any assumptions you made while reaching at your conclusion.
An automatic coffee filling process is in control when the average amount of instant coffee that is poured into a jar is 8 oz and the standard deviation is 0.2 oz. A simple random sample of 100 jars is selected and the sample average is found to be X = 8.1 Oz. Assuming the standard deviation of the process is still 0.2 oz, is the process out of control (pouring too much coffee into the jars) after specifying the following: (i) State your null and alternative hypotheses. (ii) State your level of significance. (iii) Specify any assumptions you made while reaching at your conclusion.
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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Transcribed Image Text:An automatic coffee filling process is in control when the average amount of instant coffee that is poured into a jar is
8 oz and the standard deviation is 0.2 oz. A simple random sample of 100 jars is selected and the sample average is found
to be x = 8.1 oz. Assuming the standard deviation of the process is still 0.2 oz, is the process out of control (pouring too
much coffee into the jars) after specifying the following:
(i) State your null and alternative hypotheses. (ii) State your level of significance.
(iii) Specify any assumptions you made while reaching at your conclusion.
The following`answers" have been proposed.
(a) (i) Ho : µ < 8, versus H1 :µ > 8.
(ii) We will use a = 0.05.
(iii) We will assume that the fill amount is normally distributed. We will reject Ho if Zstat > 1.645. In our case
Хn —8
Zstat =
= 5.0 > 1.645.
ol n
So, the evidence suggests that Họ should be rejected.
(b) (i) Ho : µ < 8, versus H1 :µ > 8.
(ii) We will use a = 0.01.
(iii) We will assume that the fill amount is normally distributed. We will reject Ho if Zstat > 1.645. In our case
Xn – 8
Zstat =
= 5.0 > 1.645.
ol yn
So, the evidence suggests that Họ should be rejected.
(c) (i) Ho : µ < 8, versus H1 :4 > 8.
(ii) We will use a = 0.001.
(iii) We will assume that the fill amount is normally distributed. We will reject Ho if Zstat > 1.96. In our case
Xn- 8
Zstat =
= 5.0 < 1.96.
ol n
So, the evidence suggests that Họ should not be rejected.
(d) (i) Ho : µ < 8, versus H1 : µ > 8.
(ii) We will use a = 0.05.
(iii) We will assume that the fill amount is normally distributed. We will reject Ho if Zstat > 0.05. In our case
Xn - 8
= 5.0 > 0.05.
Zstat =
ol yn
So, the evidence suggests that Họ should be rejected.
(e) None of the above.
The correct answer is
(а)
(b)
(c)
(d)
(e)
N/A
(Select One)
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