An article reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and 212 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.)

Given : An article reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and 212 of these passed the probe.

i.e.Sample size : n= 356

Sample proportion of all dies that pass the probe: [tex]\hat{p}=\dfrac{212}{356}\approx0.596[/tex]

Critical value for 95% confidence interval ( Using z-value table) :

[tex]z=1.96[/tex]

Now, the 95% (two-sided) confidence interval for the proportion of all dies that pass the probe will be :

[tex]\hat{p}\pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

i.e. [tex]0.596\pm (1.96)\sqrt{\dfrac{0.596(1-0.596)}{356}}[/tex]

i.e. [tex]0.596\pm (1.96)(0.026)[/tex]

[tex]\approx0.596\pm 0.051=(0.596-0.051,\ 0.596+0.051)\\\\=(0.545,\ 0.647)[/tex]

Hence, the 95% (two-sided) confidence interval for the proportion of all dies that pass the probe.= (0.545, 0.647)

abiolataiwo2015 abiolataiwo2015 · Answer 2 · 2022-04-12T15:34:50+02:00

The 95% (two-sided) confidence interval for the proportion of all dies that pass the probe is: 0.647; 0.545.

Confidence interval

Sample proportion = 212/356

Sample proportion =0.596

z-score for 95% confidence interval=1.96

Confidence interval=√p(1-p)/n

Confidence interval=0.596±(1.96)√0.596(1-0.596)/356

Confidence interval=0.596±(1.96)√0.596(0.404)/356

Confidence interval=0.596±(1.96)√0.240784/356

Confidence interval=0.596±(1.96)√0.00067635955

Confidence interval=0.596±(1.96) 0.026

Confidence interval=(0.596+0.051); (0.596-0.051)

Confidence interval=0.647; 0.545

Inconclusion the 95% (two-sided) confidence interval for the proportion of all dies that pass the probe is: 0.647; 0.545.

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