The quality department at ElectroTech is examining which of two microscope brands (Brand A or Brand B) to purchase. They have hired someone to inspect six circuit boards using both microscopes. Below are the results in terms of the number of defects (e.g., solder voids, misaligned components) found using each microscope. Use Table 2.
|Circuit Board||Number of defects with Brand A||Number of defects with Brand B|
|Let the difference be defined as the number of defects with Brand A – Brand B.|
Specify the null and alternative hypotheses to test for differences in the defects found between the microscope brands.
At the 5% significance level, find the critical value(s) of the test. What is the decision rule? (Negative values should be indicated by a minus sign. Round your answer to 3 decimal places.)
|(Click to select)RejectDo not reject H0 if tdf > or tdf < .|
Assuming that the difference in defects is normally distributed, calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.)
Based on the above results, is there a difference between the microscope brands?
We (Click to select) (reject/ do not reject) H0. We (Click to select ()can/cannot) conclude the mean difference between Brand A number of defects and the Brand B number of defects is different from zero.
The table given below ,
|Circuit Board||Number of defects with Brand A||Number of defects with Brand B||di=A-B||di^2|
From table ,
a. Hypothesis : Vs
b. Critical value : ; From excel "=TINV(0.05,5)
c. The value of the test statistic is ,
d. Decision : Here ,
Therefore , Do not reject Ho
Conclusion : Hence , We cannot conclude the mean difference between Brand A number of defects and the Brand B number of defects is different from zero.
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