192k views
1 vote
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

1 Answer

5 votes

Answer:

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).

User Takje
by
7.5k points