A company manufacturing computer chips finds that 8% of all chips manufactured are defective. In an effort to decrease the percentage of defective chips, management decides to provide additional training to those employees hired within the last year. After training was implemented, a sample of 450 chips revealed only 27 defects. A hypothesis test is performed to determine if the additional training was effective in lowering the defect rate. Which of the following statement is true about this hypothesis test? a) The additional training significantly increased the defect rate. b) The additional training significantly lowered the defect rate. c) The additional training did affect the defect rate. d) The additional training did not significantly lower the defect rate. e) None of these.

Question

Mathematics

Justice Kirby

24 July, 03:37

A company manufacturing computer chips finds that 8% of all chips manufactured are defective. In an effort to decrease the percentage of defective chips, management decides to provide additional training to those employees hired within the last year. After training was implemented, a sample of 450 chips revealed only 27 defects. A hypothesis test is performed to determine if the additional training was effective in lowering the defect rate. Which of the following statement is true about this hypothesis test? a) The additional training significantly increased the defect rate. b) The additional training significantly lowered the defect rate. c) The additional training did affect the defect rate. d) The additional training did not significantly lower the defect rate. e) None of these.

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Answers (1)

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Sasha Richmond · Answer 1

d) The additional training did not significantly lower the defect rate

Step-by-step 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