Question

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.

Answer 1

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

Explanation:

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

`\alpha = (1 - 0.95)/(2) = 0.025`

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

That is z with a pvalue of
`1 - 0.025 = 0.975`, so Z = 1.96.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 1.96(2)/(√(100)) = 0.392`

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.