Question

Peter is the owner of a hardware supply shop. He has hired you to check his machines calibration prior to starting production on a large order. To check this, you set the machine to create 2.0 inch nails and manufacture a random sample of 200 nails. That sample of nails has an average length of 2.024 inches with a standard deviation of 0.205 inches.


Does this sample provide convincing evidence that the machine is working properly or should it be shut down for repairs?

Answer 1

Answer:If we compare the p value and the significance level assumed we see thatso we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can conclude that the true mean is not different from the specification at 1% of signficance.  

Step-by-step explanation:

Answer 2

The machine is working properly based on the convincing evidence provided by the sample.

How to calculate the test statistics.

For the null hypothesis, we would test that:

H0 : p = 2

For the alternate hypothesis, we would test that:

Ha > p = 2

In Mathematics, the test statistics of a given sample is calculated by using this formula:

$Z=\frac{x\;-\;u}{\frac{\sigma}{\sqrt{n} } }$

Where:

• x is the sample mean.

• u is the mean.

• is the standard deviation.

• n is the sample size.

Substituting the given parameters into the formula, we have;

$Z=\frac{2.024\;-\;2}{\frac{0.205}{\sqrt{200} } }\\\\Z=\frac{0.024}{\frac{0.205}{14.14} }\\\\Z=\frac{0.024}{0.0145 }$

Z = 1.66.

For the p-value:

P(Z > Zo) = P(Z > 1.66)

P(Z > 1.66) = P(Z < -1.66)

P(Z < -1.66) = 0.0485.

From the z-table, a z-score of 1.66 corresponds has a p-value of 0.0485. Therefore, the p-value for this test is 0.0485 < 2 while assuming a confidence level of 95%.

In conclusion, we fail to reject the null hypothesis because 0.0485 < 2.

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