Question

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. Find the critical value that corresponds to a confidence level of 99%.

Answer 1

Answer:

The critical value that corresponds to a confidence level of 99% is, 2.58.

Step-by-step explanation:

Consider a random variable X that follows a Binomial distribution with parameters, sample size n and probability of success p.

It is provided that the distribution of proportion of random variable X, $\hat p$, can be approximated by the Normal distribution.

The mean of the distribution of proportion is, $\mu_{\hat p}=\hat p$

The standard deviation of the distribution of proportion is, $\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}$.

Then the confidence interval for the population proportion p is:

$CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }$

The confidence level is 99%.

The significance level is:

$\alpha =1-\frac{Confidence\ level}{100}=1-\frac{99}{100}=1-0.99=0.01$

Compute the critical value as follows:

$z_{\alpha /2}=z_{0.01/2}=z_{0.005}$

That is:

$P(Z>z)=0.005\\P(Z$

Use the z-table for the z-value.

For z = 2.58 the P (Z < z) = 0.995.

And for z = -2.58 the P (Z > z) = 0.005.

Thus, the critical value is, 2.58.