Chapter 8.4, Problem 8E

To determine

To Construct:

A 95% confidence interval for the proportion of all computer chips from one factory that are defective given that out of a random sample of 200 computer chips is obtained from the factory and 4% are found to be defective, and interpret the same.

Answer to Problem 8E

Solution:

The 95% confidence interval for the proportion of all computer chips from the factory that are defective is $\left(0.013,0.067\right)$, which means that with 95% confidence, the proportion of computer chips from the factory that are defective is between $0.013$ and $0.067$.

Explanation of Solution

Given:

Sample size $n=200$.

Individual matching the given characteristic $x=4\%\text{of}n$.

Confidence level $c=95\%$.

Approach:

A random experiment from a population of size $N$ has a fixed discrete number of trials $\left(n\right)$ out of which a certain number of them $\left(x\right)$ are regarded as successes, having a certain characteristic.

Based on these numbers, the sample proportion $\left(\widehat{p}\right)$ is obtained using the relation below:

$\widehat{p}=\frac{x}{n}$.

The sample proportions are binomially distributed (and hence are discrete), which can be approximated using a normal distribution using the continuity correction given that the sample size is large enough $\left(n\ge 30\right)$ to ensure that $n\widehat{p}\ge 5$ and $n\left(1-\widehat{p}\right)\ge 5$.

The population proportion is best estimated point-wise by the sample proportion.

The margin of error is defined as the maximum distance from point estimate that the confidence interval covers.

In terms of appropriate $z$-value and sample proportion, the margin of error is defined as:

$E=z_{\frac{\alpha }{2}}\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}$

Where $z_{\frac{\alpha }{2}}$ is the critical value whose right has the area $\frac{\alpha }{2}$, $\widehat{p}$ is the sample proportion and $n$ is the sample size.

The confidence interval bounds are equidistant from the best point estimate by the amount of margin of error as:

$\text{confidence interval}=\left(\widehat{p}-E,\widehat{p}+E\right)$.

The number of individuals in the sample after the characteristic in the question is divided by the sample size to obtain the sample proportion which is best point estimate to the population proportion.

Then, the margin of error is computed following which, the bounds of interval estimate are obtained from the best point estimate by subtracting and adding respectively margin of error.

Calculation:

The value of $x$ is obtained as:

$\begin{array}{rll}x& =& 4\%\text{of}200\\ & =& \frac{4}{100}\times 200\\ & =& 8\end{array}$

Dividing $x$ by $n$ gives:

$\begin{array}{rllll}\widehat{p}& =& \frac{x}{n}& =& \frac{8}{200}\\ & =& 0.04\end{array}$

Which gives the population proportion estimate.

Since $c=0.95$, the value of $\frac{\alpha }{2}$ is:

$\begin{array}{rll}\frac{\alpha }{2}& =& \frac{1-0.95}{2}\\ & =& \frac{0.05}{2}\\ & =& 0.025\end{array}$

The critical value is looked up in the standard normal table which yields:

$z_{0.025}=1.96$.

Then the margin of error is:

$E=1.96\times \sqrt{\frac{0.04\times \left(1-0.04\right)}{200}}$

$=1.96\times \sqrt{\frac{0.0384}{200}}$

$=1.96\times \sqrt{0.000192}$

$=1.96\times \mathrm{0.01385641...}$

$\approx 0.027159$

Rounding off to the $4^{\text{th}}$ decimal place gives the margin of error to be:

$E=0.0272$.

Then, lower bound of confidence interval is:

$\begin{array}{rll}\widehat{p}-E& =& 0.04-0.0272\\ & =& 0.0128\end{array}$

Rounding off which to the $3^{\text{rd}}$ decimal place gives $0.013$.

Then, higher bound of confidence interval is:

$\begin{array}{rll}\widehat{p}+E& =& 0.04+0.0272\\ & =& 0.0672\end{array}$

Rounding off which to the $3^{\text{rd}}$ decimal place gives $0.067$.

Thus, confidence interval is:

$\left(0.013,0.067\right)$.

Conclusion:

The 95% confidence interval for the proportion of all computer chips from the factory that are defective is $\left(0.013,0.067\right)$, which means that with 95% confidence, the proportion of computer chips from the factory that are defective is between $0.013$ and $0.067$.