Suppose that administrators at a large urban high school want to gain a better u

Question

Suppose that administrators at a large urban high school want to gain a better understanding of the prevalence of bullying within their school. They select a random sample of 250 students who are asked to complete a survey anonymously. Of these students, 89 report that they have experienced bullying. Use this information to find a 90% z-confidence interval for p, the true proportion of students at this school who have experienced bullying. Give the limits (bounds) of the confidence interval as a proportion, precise to three decimal places lower limit = upper limit =

Explanation / Answer

Solution:

Formula for confidence interval for population proportion is given as below:

Confidence interval = P -/+ Z*sqrt(P*(1 – P)/N)

We are given

Sample size = N = 250

Number of successes = X = 89

Confidence level = 90%

Sample proportion = P = X/N = 89/250 = 0.356

Critical Z value = 1.6449 (by using z-table)

Confidence interval = 0.356 -/+ 1.6449*sqrt(0.356*( 1 - 0.356)/250)

Confidence interval = 0.356 -/+ 1.6449* 0.0303

Confidence interval = 0.356 -/+ 0.049812

Lower limit = 0.356 - 0.049812 = 0.306188

Upper limit = 0.356 + 0.049812 = 0.405812

Lower limit = 0.306

Upper limit = 0.406

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