Question

An article reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and 210 of these passed the probe. Assuming a stable process, calculate a 95% (two-sided) confidence interval for the proportion of all dies that pass the probe. (Round your answers to three decimal places.

Answer 1

Concepts and reason

The confidence interval is defined as the range of values with a specified probability that the parameters lies within it.

When the procedure is repeated with more number of samples and the confidence interval is constructed at $\alpha$ level of significance, with $100\left( {1 - \alpha } \right)\%$ probability the true population parameter lies within the range of confidence interval values.

The standard error is the standard deviation of the sampling distribution of a large population.

The margin of error is defined as the half the width of the confidence interval.

A Z-score indicates the number of standard deviations an element is deviated from the mean. Z-scores may be positive or negative. The positive value indicates that the score is above the mean value. The negative value indicates that the score is below the mean value. The normal probability values can be determined with the help of Z-score.

Fundamentals

The $100\left( {1 - \alpha } \right)\%$ confidence interval for true population proportion is calculated as,

$\hat p - {z_{\alpha /2}}\sqrt {\frac{{\hat p\left( {1 - \hat p} \right)}}{n}} \le p \le \hat p + {z_{\alpha /2}}\sqrt {\frac{{\hat p\left( {1 - \hat p} \right)}}{n}}$

Here, $\hat p$ is the sample proportion and n is the sample size.

The formula for standard error is,

$SE = \sqrt {\frac{{\hat p\left( {1 - \hat p} \right)}}{n}}$

The formula for margin of error is,

$ME = {z_{\alpha /2}}\sqrt {\frac{{\hat p\left( {1 - \hat p} \right)}}{n}}$

Procedure for finding the z-value is listed below:

1.From the table of standard normal distribution, locate the probability value.

2.Move left until the first column is reached.

3.Move upward until the top row is reached.

4.The corresponding row and column values together give the value of z.

Also, the probability can be obtained from the z-table.

From the 356 dies examined by an inspection probe, 210 of them passed the probe.

The proportion of dies that passed the probe is,

$\begin{array}{c}\\\hat p = \frac{X}{n}\\\\ = \frac{{210}}{{356}}\\\\ = 0.59\\\end{array}$

The standard error of the proportion is,

$\begin{array}{c}\\SE\left( {\hat p} \right) = \sqrt {\frac{{\hat p\left( {1 - \hat p} \right)}}{n}} \\\\ = \sqrt {\frac{{0.59\left( {1 - 0.59} \right)}}{{356}}} \\\\ = \sqrt {0.00068} \\\\ = 0.0261\\\end{array}$

Confidence level, $1 - \alpha = 0.95$

Implies, level of significance, $\alpha = 0.05$

From the z-table, the critical value at $\alpha = 0.05$ for a two-tailed z-distribution is, ${z_{\alpha /2}} = 1.96$

Thus, at 95% confidence level, the critical value is 1.96.

The margin of error at 95% confidence of the proportion is,

$\begin{array}{c}\\ME\left( {\hat p} \right) = {z_{\alpha /2}} \times SE\left( {\hat p} \right)\\\\ = {z_{0.05/2}} \times 0.0261\\\\ = 1.96 \times 0.0261\\\\ = 0.051\\\end{array}$

The 95% confidence interval for the proportion of all dies that pass the probe is,

$\begin{array}{c}\\\hat p - {z_{\alpha /2}}\sqrt {\frac{{\hat p\left( {1 - \hat p} \right)}}{n}} \le p \le \hat p + {z_{\alpha /2}}\sqrt {\frac{{\hat p\left( {1 - \hat p} \right)}}{n}} \\\\\hat p - ME\left( {\hat p} \right) \le p \le \hat p + ME\left( {\hat p} \right)\\\\0.59 - 0.051 \le p \le 0.59 + 0.051\\\\0.539 \le p \le 0.641\\\end{array}$

Ans:

The 95% confidence interval for the population proportion is $\left( {0.539,\,\,0.641} \right)$ .