Answer:
d) The additional training did not significantly lower the defect rate
Explanation:
Let proportion of defective chips be = x
Null Hypothesis [H0] : Additional training has no impact on defect rate x = 8% = 0.08
Alternate Hypothesis [H1] : Additional training has impact on defect rate x < 8% , x < 0.08
Observed x proportion (mean) : x' = 27 / 450 = 0.06
z statistic = [ x' - x ] / √ [ { x ( 1-x ) } / n ]
( 0.06 - 0.08 ) / √ [ 0.08 (0.92) / 450 ]
= -0.02 / √ 0.0001635
= -0.02 / 0.01278
z = - 1.56
Since calculated value of z, 1.56 < tabulated value of z at assumed 0.01 significance level, 2.33
Null Hypothesis is accepted, 'training didn't have defect rate reduction impact' is concluded