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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

How many students in this school?

 Apr 30, 2018

I’m not sure everything i wrote is all that was provided :/

Guest Apr 30, 2018

Yeah. I can't figure out the problem with out the original students.

x = original students


x * 0.07 = ?(Percent of students error)

?(above) + x = your answer


...I think...



Mikeyy  Apr 30, 2018
edited by Mikeyy  Apr 30, 2018



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


Additional information:

For this question, the population is assumed to be above 20,000 (because no population is specified), well above what would be needed for a (US) high school population that averages 860 students per school, with the highest at 1500. For schools with these populations, the sample minimums calculate to 145 and 154 persons respectively. At 171 samples, the margin of error for these schools drops to 6.25% and 6.50% respectively.   




 May 1, 2018

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