A quality control engineer is in charge of the manufacture of flash drives. Two
ID: 2906714 • Letter: A
Question
A quality control engineer is in charge of the manufacture of flash drives. Two different processes can be used to manufacture the flash drives. He suspects that Method A produces a greater proportion of defects than Method B. He samples 150 of the Method A and 200 of the Method B flash drives and finds that 27 and 18 of them, respectively, are defective. If Method A drives are designated as'Group 1' and Method B drives are designated as 'Group 2,' perform the appropriate test at a level of significance of 0.01. Determine whether the following statement is true or false The same decision would be made with this test if the level of significance had been 0.05 rather than 0.01 True or False?Explanation / Answer
We want to test the proportion of defective items in method A is larger than Method B
Total no of records in method A = 150
and defective items in method A = 27
The proportion of defective items in method A = p0 = 27 / 150
= 0.18
Total no of records in method B = 200
and defective items in method B = 18
The proportion of defective items in method B = p0 = 27 / 200
= 0.09
overall proportion = p = ( 18 + 27 ) / ( 150 + 200 )
= 0.1286
Test of Hypothesis:
H0: Method A produce more defective items than more than B [ p0 > p1 ]
against,
H1: Method A produce more defective items than more than B [ p0 < p1 ]
Test Statistics:
Z = ( po - p1) / [ sqrt( p*(1-p)] * [sqrt(1/n1 + 1/ n2 )]
= 0.09 / [0.3348 * 0.1081]
= 2.4868
the p-value for this is P ( Z > 2.4868 ) = 0.9918 at 0.01 level of significance.
Decision Rule: If the p-value greater than 0.01 level of significance then we accept the null hypothesis
i.e. Method A produces the more defective items than B.
the p-value for this is P ( Z > 2.4868 ) = 0.9929 at 0.05 level of significance.
Decision Rule: If the p-value greater than 0.05 level of significance then we accept the null hypothesis
i.e. Method A produces the more defective items than B.
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