Question

In an accelerated failure test, components are operated under extreme conditions so that a substantial number will fail in a rather short time. In such a test involving two types of microchips, 580 chips manufactured by an existing process were tested, and 125 of them failed. Then, 780 chips manufactured by a new process were tested, and 130 of them failed. Find a 90% confidence interval for the difference between the proportions of failures for chips manufactured by the two processes. (Round the final answers to four decimal places.) The 90% confidence interval is

Answer 1

The 90% confidence interval is (0.0131, 0.0845).

Explanation:

Before finding the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
`\mu = p` and standard deviation
`s = \sqrt{(p(1-p))/(n)}`

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Old process:

125 out of 580, so:

`p_O = (125)/(580) = 0.2155`

`s_O = \sqrt{(0.2155*0.7845)/(580)} = 0.0171`

New process:

130 out of 780. So

`p_N = (130)/(780) = 0.1667`

`s_N = \sqrt{(0.1667*0.8333)/(780)} = 0.0133`

Distribution of the difference:

`p = p_O - p_N = 0.2155 - 0.1667 = 0.0488`

`s = √(s_O^2+s_N^2) = √(0.0171^2 + 0.0133^2) = 0.0217`

Confidence interval:

`p \pm zs`

In which

z is the z-score that has a p-value of
`1 - (\alpha)/(2)`.

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a p-value of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

The lower bound of the interval is:

`p - zs = 0.0488 - 1.645*0.0217 = 0.0131`

The upper bound of the interval is:

`p + zs = 0.0488 + 1.645*0.0217 = 0.0845`

The 90% confidence interval is (0.0131, 0.0845).