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# A leading medical journal claims that 68% of High school

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A leading medical journal claims that 68% of high school students experience anxiety just before an exam. A school counselor wants to conduct a survey to verify this with the students in her school and wants to do so with a margin of error (ME) of ±7%. What minimum sample size, rounded to the nearest whole person, does the counselor need to use?

N ≥ 175

N ≥ 178

N ≥ 200

N ≥ 262

Apr 30, 2018

#1
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How many students in this school?

Apr 30, 2018
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I’m not sure everything i wrote is all that was provided :/

Guest Apr 30, 2018
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Yeah. I can't figure out the problem with out the original students.

x = original students

x * 0.07 = ?(Percent of students error)

...I think...

http://www.statisticshowto.com/probability-and-statistics/hypothesis-testing/margin-of-error/

Mikeyy  Apr 30, 2018
edited by Mikeyy  Apr 30, 2018
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Solution:

$$E= Z_c \sqrt{\frac{\hat{p}(1-\hat{p})}{N}} \tiny \text{ (E is the margin of error. } Z_c \text{ is the critical value for the confidence level. } \hat{p} \text { is portion of interest.)}\\ \small \text {Arrange and solve for (N) to find the minimum sample size} \\ N= \left( \frac {Z_c /100}{E} \right)^{2} \hat{p}\left(1-\hat{p}\right) \\ \small \text {For 95% confidence }Z_c \small \text { } = 1.96\\ N= \left( \frac {1.96}{0.07} \right)^{2} \ * \ 0.68\left(1-0.68 \right) \; \ = \; \lceil{170.60}\rceil \Leftarrow \small \text {Round up to next integer}\\ \text {Minimum sample size }\mathbf {171}\\$$