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A sample of 100 fuses from a very large shipment is found to have 10 that are defective. The 95% confidence interval of defective fuses is.

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Using the confidence interval relation for one sample proportion, the confidence interval would be (0.0412; 0.1588)

The confidence interval is defined thus :

  • [p ± Z*√(p(1 - p) /n)]

Proportion, p = x/ n = 10/100 = 0.1

  • Z* at 95% confidence interval = 1.96

√(p(1 - p) /n) = [√((0.1 × 0.9) / 100)] = 0.03

  • Margin of Error = 1.96 × 0.03 = 0.0588

Lower boundary :

  • 0.1 - 0.0588 = 0.0412

Upper boundary :

  • 0.1 + 0.0588 = 0.1588

Hence, the confidence interval would be (0.0412, 0.1588)

Learn more : https://brainly.com/question/25683158

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