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Sagot :
Answer:
The p-value of the test is of 0.2776 > 0.01, which means that the we accept the null hypothesis, that is, the manager's claim that this is only a sample fluctuation and production is not really out of control.
Step-by-step explanation:
A manufacturer considers his production process to be out of control when defects exceed 3%.
At the null hypothesis, we test if the production process is in control, that is, the defective proportion is of 3% or less. So
[tex]H_0: p \leq 0.03[/tex]
At the alternate hypothesis, we test if the production process is out of control, that is, the defective proportion exceeds 3%. So
[tex]H_1: p > 0.03[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
0.03 is tested at the null hypothesis
This means that [tex]\mu = 0.03, \sigma = \sqrt{0.03*0.97}[/tex]
In a random sample of 100 items, the defect rate is 4%.
This means that [tex]n = 100, X = 0.04[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.04 - 0.03}{\frac{\sqrt{0.03*0.97}}{\sqrt{100}}}[/tex]
[tex]z = 0.59[/tex]
P-value of the test
The p-value of the test is the probability of finding a sample proportion above 0.04, which is 1 subtracted by the p-value of z = 0.59.
Looking at the z-table, z = 0.59 has a p-value of 0.7224
1 - 0.7224 = 0.2776
The p-value of the test is of 0.2776 > 0.01, which means that the we accept the null hypothesis, that is, the manager's claim that this is only a sample fluctuation and production is not really out of control.
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