Question

20. Jorge conducted a survey by asking 120 students whether they use a pencil to take notes. Of the students surveyed, 90 responded favorably. Which margin of error matches each confidence level for this situation?

1. 90% =
2. 99% =
3. 95%=
options- 0.102, 0.338, 0.216, 0.077, 0.256, 0.065

Answer 1

Answer: Margin of error for 99% CI: z=2.575 . input that into the equation and it give you .1017

Explanation:

Answer 2

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

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
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

For this problem, we have that:

120 students, so
`n = 120`

90 responded favorably, so
`p = (90)/(120) = 0.75`

Margin of error for 90% CI:

Here
`\alpha = 0.1`, z is the value of Z that has a pvalue of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)} = 1.645\sqrt{(0.75*0.25)/(120) = 0.065`

Margin of error for 99% CI:

Here
`\alpha = 0.01`, z is the value of Z that has a pvalue of
`1 - (0.01)/(2) = 0.995`, so
`Z = 2.325`.

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)} = 2.325\sqrt{(0.75*0.25)/(120) = 0.091`

Margin of error for 95% CI:

Here
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)} = 1.96\sqrt{(0.75*0.25)/(120) = 0.077`