Question

Jonah wants to construct a confidence interval using 90% confidence to estimate what proportion of silicon wafers at his factory is defective. He wants the margin of error to be no more than 3%. A previous study suggests that about 6% of these wafers are defective. If we assume p=0.06, what is the smallest sample size required to obtain the desired margin of error? 234 416 170 936

Answer 1

The smallest sample size required to obtain the desired margin of error is 936.

Step-by-step explanation:

To determine the smallest sample size required to obtain the desired margin of error, we can use the formula:

n = (Z^2 * p * (1-p)) / E^2

where:

n is the required sample size

Z is the Z-score corresponding to the desired confidence level (in this case, 90%)

p is the estimated proportion of defective silicon wafers (0.06)

E is the margin of error (0.03)

Substituting the values into the formula, we get:

n = ((1.645)^2 * 0.06 * (1-0.06)) / (0.03)^2

Simplifying the equation, we find that the smallest sample size required is 936.

Answer 2

170

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is of:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

Assume:

`\pi = 0.06`

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a pvalue of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

What is the smallest sample size required to obtain the desired margin of error?

This is n for which M = 0.03. So

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.03 = 1.645\sqrt{(0.06*0.94)/(n)}`

`0.03√(n) = 1.645√(0.06*0.94)`

`√(n) = (1.645√(0.06*0.94))/(0.03)`

`(√(n))^2 = ((1.645√(0.06*0.94))/(0.03))^2`

`n = 169.6`

Rounding up, 170.