Answer :
Answer:
The correct answer is
(0.0128, 0.0532)
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 interval [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 zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex]
For this problem, we have that:
In a random sample of 300 circuits, 10 are defective. This means that [tex]n = 300[/tex] and [tex]\pi = \frac{10}{300} = 0.033[/tex]
Calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool.
So [tex]\alpha[/tex] = 0.05, z is the value of Z that has a pvalue 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)}{300}} = 0.033 - 1.96\sqrt{\frac{0.033*0.967}{300}} = 0.0128[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{300}} = 0.033 + 1.96\sqrt{\frac{0.033*0.967}{300}} = 0.0532[/tex]
The correct answer is
(0.0128, 0.0532)