Respuesta :
Answer:
A sample of 1032 is needed.
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 zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
A sample of 300 components showed that 20 were defective.
This means that [tex]\pi = \frac{20}{300} = 0.0667[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
How large a sample is needed to estimate the true proportion of defective components to within 2.5 percentage points with 99% confidence?
A sample of n is needed.
n is found when M = 0.025. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.025 = 2.575\sqrt{\frac{0.0667*0.9333}{n}}[/tex]
[tex]0.02\sqrt{n} = 2.575\sqrt{0.0667*0.9333}[/tex]
[tex]\sqrt{n} = \frac{2.575\sqrt{0.0667*0.9333}}{0.02}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.575\sqrt{0.0667*0.9333}}{0.02})^2[/tex]
[tex]n = 1031.9[/tex]
Rounding up
A sample of 1032 is needed.
