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Scott is the owner of a tool and die shop. He has hired you to check his machine’s calibration prior to starting production on a large order. To check this, you set the machine to create 1.5 inch screws and manufacture a random sample of 200 screws. That sample of screws has an average length of 1.476 inches with a standard deviation of 0.203 inches.

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

Sagot :

Using the t-distribution, it is found that we do not reject the null hypothesis, which is convincing evidence that the machine is working properly.

What are the hypothesis tested?

At the null hypothesis, it is tested if the mean is of 1.5 inches, that is:

[tex]H_0: \mu = 1.5[/tex].

At the alternative hypothesis, it is tested if it is different, hence:

[tex]H_1: \mu \neq 1.5[/tex].

What is the test statistic?

The test statistic is given by:

[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]

The parameters are:

  • [tex]\overline{x}[/tex] is the sample mean.
  • [tex]\mu[/tex] is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

In this problem, their values are:

[tex]\overline{x} = 1.476, \mu = 1.5, s = 0.203, n = 200[/tex].

Hence, the test statistic is given by:

[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]

[tex]t = \frac{1.476 - 1.5}{\frac{0.203}{\sqrt{200}}}[/tex]

[tex]t = -1.672[/tex]

What is the decision?

Considering a two-tailed test, as we are testing if the mean is diffrent of a value, with 200 - 1 = 199 df and a standard significance level of 0.05, the critical value is of [tex]|t^{\ast} = 1.972[/tex].

Since the absolute value of the test statistic is less than the critical value, we do not reject the null hypothesis, which is convincing evidence that the machine is working properly.

More can be learned about the t-distribution at https://brainly.com/question/16313918

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