SOLUTION: A sample of 400 nursing applications included 220 from women. Find the 95% confidence interval of the true proportion of women who applied to the nursing program.
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Question 1062600: A sample of 400 nursing applications included 220 from women. Find the 95% confidence interval of the true proportion of women who applied to the nursing program. Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! sample proportion(p) is 220/400 = 0.55
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standard error(se) = sqrt( p * (1-p) / n ) = sqrt ( 0.55 * 0.45 / 400 ) = 0.0249
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Note that n is the sample size = 400
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margin of error(me) = critical value * se
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our confidence interval is 95%
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alpha(a) = 1 - (95/100) = 0.05
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critical probability(p*) = 1 - (0.05/2) = 0.975
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We use the normal distribution(since n > 30) to determine the critical value as a z-score, find the z-score having a cumulative probability equal to the critical probability (p*).
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cv = 1.96
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me = 1.96 * 0.0249 = 0.0488
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Our 95% confidence interval is
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0.55 + or - 0.0488
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