Question 317193: A survey of an urban university (population of 25,450) showed that 870 of 1100 students sampled supported a fee increase to fund improvements to the student recreation center
Using the 95% level of confidence, what is the confidence interval?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A survey of an urban university (population of 25,450) showed that 870 of 1100 students sampled supported a fee increase to fund improvements to the student recreation center
Using the 95% level of confidence, what is the confidence interval?
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sample proportion: 870/1100 = 0.7909
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ME = 1.96*sqrt[0.79*0.21/1100] = 0.0241
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95% CI: 0.79 - 0.0241 < p < 0.79 + 0.0241
95% CI: 0.7659 < p < 0.8141
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Cheers,
Stan H.
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