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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
35 random patient files and finds that 27 of these patients are using this same toothpaste.
This means that [tex]n = 35, \pi = \frac{27}{35} = 0.7714[/tex]
a. What is the margin of error for a 95% confidence interval to the nearest tenth of a percent?
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]M = 1.96\sqrt{\frac{0.7714*0.2286}{35}}[/tex]
[tex]M = 0.139[/tex]
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?
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
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.
Answer:
a) 13.9%.
b) If the number of patient files in this study were increased to 100, the margin of error would be decreased.
Step-by-step explanation:
