Question

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. The correct value of the Z-statistic is

Answer 1

Answer:

The correct value of the Z-statistic is z = -1.56

Step-by-step explanation:

A company manufacturing computer chips finds that 8% of all chips manufactured are defective.

This means that the null hypothesis is:

$H_{0}: p = 0.08$

A hypothesis test is performed to determine if the additional training was effective in lowering the defect rate.

This means that the alternate hypothesis is:

$H_{a}: p < 0.08$

z-statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.08 is tested at the null hypothesis:

This means that $\mu = 0.08, \sigma = \sqrt{0.08*0.92}$

After training was implemented, a sample of 450 chips revealed only 27 defects.

This means that $n = 450, X = \frac{27}{450} = 0.06$

The correct value of the Z-statistic is

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.06 - 0.08}{\frac{\sqrt{0.08*0.92}}{\sqrt{450}}}$

$z = -1.56$