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Stainless steels can be susceptible to stress corrosion cracking under certain conditions. A materials engineer is interested in determining the proportion of steel alloy failures that are due to stress corrosion cracking. In the absence of preliminary data, what is the minimum sample size required to ensure a 90% confidence interval for p to within 0.02

Respuesta :

Using the z-distribution, as we are working with a proportion, the minimum sample size required to ensure a 90% confidence interval for p to within 0.02 is of 1692.

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which:

  • [tex]\pi[/tex] is the sample proportion.
  • z is the critical value.
  • n is the sample size.

The margin of error is of:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In this problem:

  • No prior estimate, hence the proportion is [tex]\pi = 0.5[/tex].
  • Margin of error of M = 0.02.
  • 90% confidence level, hence[tex]\alpha = 0.9[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.9}{2} = 0.95[/tex], so [tex]z = 1.645[/tex].

Then, solving for n:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.02 = 1.645\sqrt{\frac{0.5(0.5)}{n}}[/tex]

[tex]0.02\sqrt{n} = 1.645(0.5)[/tex]

[tex]\sqrt{n} = \frac{1.645(0.5)}{0.02}[/tex]

[tex](\sqrt{n})^2 = \left(\frac{1.645(0.5)}{0.02}\right)^2[/tex]

[tex]n = 1691.3[/tex]

Rounding up, a sample size of 1692 is needed.

More can be learned about the z-distribution at https://brainly.com/question/25890103