You intend to estimate a population proportion with a confidence interval. The data suggests that the normal distribution is a reasonable approximation for the binomial distribution in this case. While it is an uncommon confidence level, find the critical value that corresponds to a confidence level of 97.1%.

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Answer:

The critical value that corresponds to a confidence level of 97.1% is [tex]Z = 2.18[/tex].

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 z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].

97.1% confidence level

So [tex]\alpha = 0.029[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.029}{2} = 0.9855[/tex], so [tex]Z = 2.18[/tex].

The critical value that corresponds to a confidence level of 97.1% is [tex]Z = 2.18[/tex].