Answer:
Margin of error for 90% C.I: 0.065
Margin of error for 99% C.I: 0.091
Margin of error for 95% C.I: 0.077
Step-by-step explanation:
In this question, for each student, there are only two possible outcomes. Either they use a pencil to take notes, or they do not. Students are independent of each other. So we use the binomial probability distribution to solve this question.
Binomial confidence interval:
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]
For this problem, we have that:
120 students, so [tex]n = 120[/tex]
90 responded favorably, so [tex]p = \frac{90}{120} = 0.75[/tex]
Margin of error for 90% CI:
Here [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}} = 1.645\sqrt{\frac{0.75*0.25}{120} = 0.065[/tex]
Margin of error for 99% CI:
Here [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.325[/tex].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}} = 2.325\sqrt{\frac{0.75*0.25}{120} = 0.091[/tex]
Margin of error for 95% CI:
Here [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].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}} = 1.96\sqrt{\frac{0.75*0.25}{120} = 0.077[/tex]
