A manufacturer must test that his bolts are 4.00 cm long when they come off the assembly line. He must recalibrate his machines if the bolts are too long or too short. After sampling 196 randomly selected bolts off the assembly line, he calculates the sample mean to be 4.14 cm. He knows that the population standard deviation is 0.83 cm. Assuming a level of significance of 0.05, is there sufficient evidence to show that the manufacturer needs to recalibrate the machines? State the null and alternative hypotheses for the test.

Answer:

The null hypothesis is [tex]H_o: \mu = 4[/tex].

The alternate hypothesis is [tex]H_a: \mu \neq 4[/tex]

The pvalue of the test is 0.0182 < 0.05, which means that there is sufficient evidence to show that the manufacturer needs to recalibrate the machines.

Step-by-step explanation:

A manufacturer must test that his bolts are 4.00 cm long when they come off the assembly line.

This means that the null hypothesis is that they are 4.00 cm long, that is:

[tex]H_o: \mu = 4[/tex]

At the alternate hypothesis we test if it is different. So

[tex]H_a: \mu \neq 4[/tex]

The test statistic is:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

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

4 is tested at the null hypothesis:

This means that [tex]\mu = 4[/tex]

After sampling 196 randomly selected bolts off the assembly line, he calculates the sample mean to be 4.14 cm.

This means that [tex]n = 196, X = 4.14[/tex]

He knows that the population standard deviation is 0.83 cm.

This means that [tex]\sigma = 0.83[/tex]

Value of the test statistic:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{4.14 - 4}{\frac{0.83}{\sqrt{196}}}[/tex]

[tex]z = 2.36[/tex]

Pvalue of the test and decision:

The pvalue of the test is the probability that the sample mean is different by at least 4.14 - 4 = 0.14 from the target value, which is P(|z| > 2.36), which is 2 multiplied by the pvalue of z = -2.36.

Looking at the z-table, z = -2.36 has a pvalue of 0.0091

2*0.0091 = 0.0182

The pvalue of the test is 0.0182 < 0.05, which means that there is sufficient evidence to show that the manufacturer needs to recalibrate the machines.

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