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

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

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{(\hat p(1-\hat p))/(n)}`.

Then the confidence interval for the population proportion p is:

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

The confidence level is 99%.

The significance level is:

`\alpha =1-(Confidence\ level)/(100)=1-(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<z)=1-0.005 = 0.995`

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.