There is a population of millions of computer chips produced at a factory. The production process is so precious that it is known that the population standard deviation for their mass is 2 milligrams. We want to get a 95% confidence interval for the mass of a single microchip produced here. So we randomly sample 100 microchips. The sample average mass turns out to be 1972 milligrams. They found this by placing them all on a scale at once and finding a mass of 197.2 grams, and then dividing by 100. You don't think they would actually weigh each one individually, do you? That would be super annoying.

Required:
Find a 95% confidence interval for the mass of a microchip produced at this factory.

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joaobezerra joaobezerra · Answer 1 · 2021-05-31T20:28:13+02:00

Answer:

The 95% confidence interval for the mass of a microchip produced at this factory is between 1971.608 milligrams and 1972.392 milligrams.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]

Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 1.96\frac{2}{\sqrt{100}} = 0.392[/tex]

The lower end of the interval is the sample mean subtracted by M. So it is 1972 - 0.392 = 1971.608 milligrams

The upper end of the interval is the sample mean added to M. So it is 1972 + 0.392 = 1972.392 milligrams

The 95% confidence interval for the mass of a microchip produced at this factory is between 1971.608 milligrams and 1972.392 milligrams.