Question

Elena wants to build a one-sample z interval to estimate what proportion of computers produced at a

factory have a certain defect. She chooses a confidence level of 94%. A random sample of 200 computers

shows that 12 computers have the defect

What critical value z' should Elena use to construct this confidence interval?

Answer 1

Answer:

Z = 1.88

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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

94% confidence level

So $\alpha = 0.06$, z is the value of Z that has a pvalue of $1 - \frac{0.06}{2} = 0.97$, so $Z = 1.88$.

Answer 2

The critical z-value elena should use to construct the given confidence interval is; 1.88

We are given;

• Confidence level = 94%
• Sample proportion; p^ = 12/200 = 0.06
• Sample size; n = 200

Formula for confidence intervals of proportion is;

CI = p^ ± z√(p^(1 - p^)/n)

Since we are given Confidence level of 94%, the significance level is;

α = 100% - 94%

α = 0.06

The the z-score will be at the p-value of 1 - α/2

P-value = 1 - (0.06/2) = 0.97

From online p-value from z-score calculator, we have; z-score = 1.88

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