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A survey was conducted to determine what percentage of college seniors would have chosen to attend a different college if they had known then what they know now. in a random sample of 100 seniors, 34 percent indicated that they would have attended a different college. create a 90 percent confidence interval for the percentage of all seniors who would have attended a different college.

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asmak525
GDC: 1 - Prop Z Int       n = 100      CL= .96     X= (100) (.34) =34     (.26208, .41792)    =26.2%tp, 41.8%

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Using the confidence interval relation for one sample proportion, the confidence interval would be (0.2926, 0.3874)

The confidence interval is defined thus :

  • [p ± Z*√(p(1 - p) /n)]

Proportion, p = 0.34

  • Z* at 90% confidence interval = 1.645

√(p(1 - p) /n) = [√((0.34 × 0.66) / 100)] = 0.03

Margin of Error = 1.96 × 0.03 = 0.0474

Lower boundary :

  • 0.34 - 0.0474 = 0.2926

Upper boundary :

  • 0.34 + 0.0474 = 0.3874

Hence, the confidence interval would be (0.2926, 0.3874)

Learn more : brainly.com/question/25683158