Question

A dentist notices that many of his patients that have been complaining of gum soreness have been using a particular brand of teeth whitening toothpaste. To determine what proportion of his patients are using this toothpaste, he selects 35 random patient files and finds that 27 of these patients are using this same toothpaste.
a. What is the margin of error for a 95% confidence interval to the nearest tenth of a percent?

b. If the number of patient files in this study were increased to 100, what would happen to the margin of error?

Answer 1

Answer:

a) 13.9%.

b) If the nuber of patient files in this study were increased to 100, the margin of error would be decreased.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

35 random patient files and finds that 27 of these patients are using this same toothpaste.

This means that $n = 35, \pi = \frac{27}{35} = 0.7714$

a. What is the margin of error for a 95% confidence interval to the nearest tenth of a percent?

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$M = 1.96\sqrt{\frac{0.7714*0.2286}{35}}$

$M = 0.139$

As a percent, a margin of error of 13.9%.

b. If the number of patient files in this study were increased to 100, what would happen to the margin of error?

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

This means that the margin of error is inversely proportional to the sample size. As n increases, n decrease. Then:

If the nuber of patient files in this study were increased to 100, the margin of error would be decreased.