Question

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.

Answer 1

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.

Answer 2

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

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 (
  `(112)/(300)=0.373`)

• ME is the margin of error from the mean

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

ME=
`(z*√(p*(1-p)))/(√(N) )` where

• z is the corresponding statistic in 98% confidence level (2.33)

• p is the sample proportion (
  `(112)/(300)=0.373`

• 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