In a simple random sample of 90 patients who saw a certain dentist, 8 patients had their teeth whitened. which interval is the 95% confidence interval for the percent of all the dentists patients who had their teeth whitened?

(3.01%, 14.77%)

Step-by-step explanation:

The confidence interval of a proportion is:

CI = p ± SE × CV,

where p is the proportion, SE is the standard error, and CV is the critical value (either a t-score or a z-score).

We already know the proportion: 8/90.  But we need to find the standard error and the critical value.

The standard error is:

SE = √(p (1-p) / n)

SE = √((8/90) * (82/90) / 90)

SE = 0.03

To find the critical value, we must first find the alpha level and the degrees of freedom.

The alpha level for a 95% confidence interval is:

α = (1 - 0.95) / 2 = 0.025

The degrees of freedom is one less than the sample size:

df = n - 1 = 90 - 1 = 89

Since df > 30, we can approximate this with a normal distribution.

If we look up the alpha level in a z score table, we find the z-score is 1.96.  That's our critical value.  CV = 1.96.

Now we can find the confidence interval:

CI = 8/90 ± 0.03 * 1.96

CI = 0.0889 ± 0.0588

CI = (0.0301, 0.1477)

So we are 95% confident that the percent of patients who had their teeth whitened is between 3.01% and 14.77%.

its 69.420

step-by-step explanation:

no this is a nonlinear negative