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, [tex]\hat p[/tex], can be approximated by the Normal distribution.
The mean of the distribution of proportion is, [tex]\mu_{\hat p}=\hat p[/tex]
The standard deviation of the distribution of proportion is, [tex]\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex].
Then the confidence interval for the population proportion p is:
[tex]CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex]
The confidence level is 99%.
The significance level is:
[tex]\alpha =1-\frac{Confidence\ level}{100}=1-\frac{99}{100}=1-0.99=0.01[/tex]
Compute the critical value as follows:
[tex]z_{\alpha /2}=z_{0.01/2}=z_{0.005}[/tex]
That is:
[tex]P(Z>z)=0.005\\P(Z<z)=1-0.005 = 0.995[/tex]
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.
