E 100 female college students in universities in the United States, 15 gestion 2

Question

E 100 female college students in universities in the United States, 15 gestion 2. In a sample of ropited as being victims of sexual assaults in the campus rted as being vitims of a) Use the most appropriate test to estimate the true proportion of female coll universities whowere victims of sexual assaults using a 99% Confidence Interval(n 5pts (b) What is the margin of error for this study? Use a-0.01 (Spts) (o) How many female college students should have been surveyed if we wanted to have a margin of error of 5%? Use =0.01. (5pts) (d) Use the sample size you computed above and test the hypothesis that more than 20% of the temale college students are victims of the sexual assaults in the campuses-compute the p-value For this question, assume that 18 596 female college students you surveyed responded positively to your question Make sure to state your null and alternative hypothesis. Use a o.01 (10pts)

Explanation / Answer

A) p = 15/100 = 0.15

This is a test of proportion.

B) At 99% confidence interval the critical value is 2.58

Margin of error = z0.005 * sqrt (p * (1 - p)/n )

= 2.58 * sqrt (0.15 * 0.85/100)

= 0.092

C) margin of error = 0.05

Or, z0.005 * sqrt (p * (1 - p)/n ) = 0.05

Or, 2.58 * sqrt (0.15 * 0.85/n) = 0.05

Or, sqrt(n) = 2.58 * sqrt(0.15 * 0.85)/0.05

Or, sqrt(n) = 18.42

Or, n = 339

D) H0: P = 0.2

H1: P > 0.2

The test statistic Z = (p - P)/sqrt (P * (1 - P)/n)

= (0.185 - 0.2)/sqrt (0.2 * 0.8/339)

= -0.69

P-value = P(Z < -0.69)

= 0.2451

As the p-value is greater than the significance level (0.2451 > 0.01), so the null hypothesis is not rejected.

So at 1% significance level there is not sufficient evidence that more than 20% of the female college students are the victims of the sexual assaults in the campus.

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