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A survey of 300 union members in New York State reveals that 112 favor the Republican candidate for governor. Construct the 98% confidence interval for the true population proportion of all New York State union members who favor the Republican candidate.

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Answer:

98% confidence interval for the true population proportion of all New York State union members who favor the Republican candidate is 0.373±0.065, that is between 0.308 and 0.438

Step-by-step explanation:

Confidence Interval can be calculated using p±ME where

  • p is the sample proportion of 300 union members in New York State who favor the Republican candidate ([tex]\frac{112}{300}=0.373[/tex])
  • ME is the margin of error from the mean

and margin of error (ME) around the mean can be found using the formula

ME=[tex]\frac{z*\sqrt{p*(1-p)}}{\sqrt{N} }[/tex] where

  • z is the corresponding statistic in 98% confidence level (2.33)
  • p is the sample proportion ([tex]\frac{112}{300}=0.373[/tex]
  • N is the sample size (300)

Then ME=[tex]\frac{2.33*\sqrt{0.373*0.627}}{\sqrt{300} }[/tex≈0.065

98% confidence interval is 0.373±0.065

Answer:

Step-by-step explanation:

Of 80 adults selected randomly from one

town, 64 have health insurance. Find a 90%

confidence interval for the true proportion of

all adults in the town who have health

insurance.

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