Answer:
[0.5935, 0.6930]
Step-by-step explanation:
The 95% confidence interval is given by
[tex]\bf p\pm t^*\sqrt{p(1-p)/n}[/tex]
where
p = the proportion of dies that passed the probe = 229/356 = 0.6432
[tex]\bf t^*[/tex] the Student's t distribution value for a 95% confidence level and 355 degrees of freedom (sample size -1)
n = sample size
Since the sample size is big enough, [tex]\bf t^*[/tex] equals the value [tex]\bf z^*[/tex] for the 95% confidence level associated with the Normal distribution N(0,1) [tex]\bf z^*[/tex] = 1.96
This value can be found either with a table or with a spreadsheet.
In Excel use NORM.INV(0.975,0,1)
In OpenOffice Calc use NORMINV(0.975;0;1)
We get a value of [tex]\bf z*[/tex]= 1.96
and our 95% confidence interval is
[tex]\bf 0.6432\pm 1.96\sqrt{0.643*0.357/356}=0.6432\pm 1.96*0.0254=0.6432\pm 0.0497[/tex] = [0.5935, 0.6930]
To interpret this result, we could say there is a 95% of probability that the proportion of dies that should pass the inspection process is between 59.35% and 69.30%
