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Murljashka [212]
2 years ago
15

a specification for bottle filling has an upper limit of 9.16 ounces and a lower limit of 9.12 ounces. the process is stabilized

and centered at the specification mean, with a process width of 0.05 ounces. what is the process capability ratio
Mathematics
1 answer:
ivanzaharov [21]2 years ago
7 0

Answer:

Process Capability ratio / Index = 0.133

Step-by-step explanation:

From the question, we are told:

Upper limit of 9.16 ounces

Lower limit of 9.12 ounces

Process width of 0.05 ounces.

what is the process capability ratio

Process Capability ratio or Index formula

= Cp = (USL - LSL)/(6 × Standard deviation)

Cp = 9.16 - 9.12/6 × 0.05

Cp = 0.04/0.3

Cp = 0.13333333333

Approximately = 0.133

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Answer:

a) Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

b) df=n-1=25-1=24  

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c) Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

Step-by-step explanation:

Data given and notation  

\bar X=18.3 represent the sample mean

s=3.72 represent the sample standard deviation

n=25 sample size  

\mu_o =15 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

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We need to conduct a hypothesis in order to check if the true mean for vacation days is higher than 15, the system of hypothesis would be:  

Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Part b: P-value  and conclusion

The first step is calculate the degrees of freedom, on this case:  

df=n-1=25-1=24  

For this case the p value is given p_v = 0.0392

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.  

Part c

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So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

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Step-by-step explanation:

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