Chapter 8.4, Problem 12E

To determine

To Construct:

A 99% confidence interval for the percentage of all computer chips from that factory that are not defective

Answer to Problem 12E

Solution:

A 99% confidence interval for the percentage of all computer chips from that factory that are not defective is $\left(92.4,99.5\right)$

Explanation of Solution

Given:

Sample size $n=200$.

Individual not matching the given characteristic $x=4\%\text{of}n$.

Confidence level $c=99\%$.

Description:

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 following 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:

$\begin{array}{rll}x& =& 200-\left(4\%\text{of}200\right)\\ & =& 200-\left(\frac{4}{100}\times 200\right)\\ & =& 200-8\\ & =& 192\end{array}$

Dividing $x$ by $n$ gives:

$\begin{array}{rllll}\widehat{p}& =& \frac{x}{n}& =& \frac{192}{200}\\ & =& 0.96\end{array}$

Which gives the population proportion estimate.

Since $c=0.99$, the value of $\frac{\alpha }{2}$ is:

$\begin{array}{rll}\frac{\alpha }{2}& =& \frac{1-0.99}{2}\\ & =& \frac{0.01}{2}\\ & =& 0.005\end{array}$

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

$z_{0.005}=2.57$.

Then the margin of error is:

$\begin{array}{rll}E& =& 2.57\times \sqrt{\frac{0.96\times \left(1-0.96\right)}{200}}\\ & =& 2.57\times \sqrt{\frac{0.0384}{200}}\\ & =& 2.57\times \sqrt{0.000192}\\ & =& 2.57\times \mathrm{0.01385641...}\\ & \approx & 0.035610964\end{array}$

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

$E=0.0356$.

Then, lower bound of confidence interval is:

$\begin{array}{rll}\widehat{p}-E& =& 0.96-0.0356\\ & =& 0.9244\end{array}$

Rounding off which to the $3^{\text{rd}}$ decimal place gives $0.924$, which converted into percentage gives $92.4\%$.

Then, higher bound of confidence interval is:

$\begin{array}{rll}\widehat{p}+E& =& 0.96+0.0356\\ & =& 0.9956\end{array}$

Rounding off which to the $3^{\text{rd}}$ decimal place gives $0.996$, which converted into percentage gives $99.5\%$.

Thus, confidence interval is:

$\left(92.4,99.5\right)$.

Conclusion:

A 99% confidence interval for the percentage of all computer chips from that factory that are not defective is $\left(92.4,99.5\right)$, which means that with 99% confidence, the percentage of all computer chips from that factory that are not defective is between $92.4\%$ and $99.5\%$.