Question #184149

A sample of 150 students were chosen at random from all grade 11 students in a private school in Pampanga. The sample indicated that 75% of them were in favor of having an educational trip. Find a) 95% and b) 99% confidence intervals for the proportion of all students who are in favor of having an educational trip. 


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Expert's answer
2021-05-07T09:05:00-0400

0.95 confidence intervalL.L=p^zα2p^(1p^)n=0.751.960.75(0.25)1500.681U.L=p^+zα2p^(1p^)n0.75+1.960.75(0.25)1500.8190.99 confidence intervalL.L=p^zα2p^(1p^)n=0.751.650.75(0.25)1500.692U.L=p^+zα2p^(1p^)n0.75+1.650.75(0.25)1500.808\text{0.95 confidence interval}\\ L.L=\hat p - z_\frac{\alpha} {2} \sqrt{\frac{\hat p (1-\hat p) }{n}} \\ =0.75 - 1.96\sqrt{\frac{0.75(0.25) }{150}} \approx 0.681\\ U.L=\hat p +z_\frac{\alpha} {2} \sqrt{\frac{\hat p (1-\hat p) }{n}} \\ 0.75 +1.96\sqrt{\frac{0.75(0.25) }{150}} \approx 0.819\\ \text{0.99 confidence interval}\\ L.L=\hat p - z_\frac{\alpha} {2} \sqrt{\frac{\hat p (1-\hat p) }{n}} \\ =0.75 - 1.65\sqrt{\frac{0.75(0.25) }{150}} \approx 0.692\\ U.L=\hat p +z_\frac{\alpha} {2} \sqrt{\frac{\hat p (1-\hat p) }{n}} \\ 0.75 +1.65\sqrt{\frac{0.75(0.25) }{150}} \approx 0.808\\


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