# 1231234

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• Published : December 23, 2012

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1. a) Proportion of all acceptable applicants who accept Harrigan’s invitation to enroll 103/178 = 0.5787 = sample mean
Binomial distribution
Mean = 0.5 = np, Standard deviation = {np(1-p)}^0.5 = (0.5*(1-0.5))^0.5 = 0.5 95% confidence intrerval
= 0.5787 ± 1.96*0.5/(178)^0.5
=0.5025~0.6521 = 50.25% ~ 65.21%
b) less than 330
33/103 = 0.3204 = sample mean
95% confidence interval
= 0.3204 ± 1.96*0.5/(103)^0.5
=0.2318~0.4195 = 23.18%~41.95%
c) between 330 and 375

d) more than 375
15/103 = 0.1456 = sample mean
95% confidence interval
= 0.1456 ± 1.96*0.5/(103)^0.5
=0.0839~0.2288 = 8.39%~22.88%
2. Find a 95% confidence interval for the proportion of all acceptable students with a combined score less than the median(356) who choose Harrigan’s rival over Harrigan. Do the same for those with a combined score greater than the median. “less than median” = 26/75 = 0.3467 = sample mean

95% confidence interval
= 0.3467 ± 1.96*0.5/(75)^0.5
= 23.35%~45.98%
“greater than median” = 48/75 = 0.64 = sample mean
95% confidence interval
= 0.64 ± 1.96*0.5/(75)^0.5
= 52.68%~75.32%
4. a) 95% confidence interval for the proportion of all students who decide to enroll at Harrigan who have been officers of at least two clubs 83/103 = 0.8058 = sample mean
95% confidence interval
= 0.8058 ± 1.96*0.5/(103)^0.5

http://speedpointer.blogspot.kr/2011/08/stat.html