American Oil Company found on pages
First issue in the case is that company is worried that whether the suppliers will be able to produce parts with its desired mean and standard deviation or not? Second issue is that how many parts should be bought by the company for enough production of two exploration devices for oil. These are the two main issues.
Chi Square test can be used for this purpose. It can be used to understand that whether the claim of the suppliers is true or not?
Null hypothesis is that given data corresponds to normal distribution
Alternate hypothesis: Given data corresponds to the normal distribution.
Sample size: =330
|0.14 and under 0.15||70||0.15||0.149882285|
|0.15 and under 0.16||90||0.16||0.191462461|
|0.16 and under 0.17||105||0.17||0.191462461|
|0.17 and under 0.18||50||0.18||0.149882285|
|0.14 and under 0.15||70||0.15||0.149882285||49||9|
|0.15 and under 0.16||90||0.16||0.191462461||63||11.57143|
|0.16 and under 0.17||105||0.17||0.191462461||63||28|
|0.17 and under 0.18||50||0.18||0.149882285||49||0.020408|
T statistics= 124
Various tests can be used to find out the statistical values. It has obtained that
α = 0.05
Critical path = χ² (0.05, 6) = 11.07
Degree of freedom= 5
Null hypothesis would be rejected based on the evidence that test statistics are greater than the critical value. The evidence has provided below:
T statistics= 124
Critical path= 11.07
Difference= 113= risky
Based on the evidences, null hypothesis should be rejected.
Two problems have discussed in this case. First is that whether the supplier would be able to provide the components having expected mean and S.D of the company or not and second is the number of components that should be purchased. After this case analysis, S.D of 0.02 with mean of 0.16 has tested and concluded that supplier would be able to produce it.