Respuesta :
Answer:
The 95% confidence interval for the proportion of all dies that pass the probe is (0.637, 0.733).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
356 dies were examined by an inspection probe and 244 of these passed the probe.
This means that [tex]n = 356, \pi = \frac{244}{356} = 0.685[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.685 - 1.96\sqrt{\frac{0.685*0.315}{356}} = 0.637[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.685 + 1.96\sqrt{\frac{0.685*0.315}{356}} = 0.733[/tex]
The 95% confidence interval for the proportion of all dies that pass the probe is (0.637, 0.733).
