The manufacturer of M&M;\'s candies states that 16% of the candies are green. Yo

Question

The manufacturer of M&M;'s candies states that 16% of the candies are green. You pick a large bag at random that contains 200 candies and find 20% of the candies are green. What's the probability of getting a sample proportion of green M&M;'s of 20% or higher? About 46% About 6% Less than 0.00001 About 80% In a recent General Social Survey, 44.3% of 1, 980 respondents said that they had a gun at home What's the margin of error for a 95% confidence interval to estimate the population proportion who have a gun at home? About plus or minus 2.2% About plus or minus 1.1% About plus or minus 0.021% About plus or minus 2.9% An NBC News/Wall Street Journal poll held the week before the 2012 presidential election found that 48% of 1, 475 likely voters said they planned to vote for Obama. What's a 90% confidence interval for the population proportion who said they planned to vote for Obama? 46.7% to 49.3% 45.9% to 50.1% 44.6% to 51.4% 45.5% to 50.5%

Explanation / Answer

27) here p=0.16 ; n=200

std error =(p(!-p)/n)1/2 =0.026

P(X>0.20)=1-P(Z<(0.20-0.16)/0.026) =1-P(Z<1.543)=1-0.9386 =0.0614

hence option about 6% is correct

28)for p=0.443 ; n=1980 ; std error =(p(!-p)/n)1/2 =0.011

for 95% CI, z=1.96

hence margin of error =z*std error =0.0218~about plus or minus 2.2%

29)

p=0.48 ; n=1475 ; std error =(p(!-p)/n)1/2 =0.013

for 95% CI, z=1.645

hence margin of error =z*std error =0.0214

hence confidence interval =sample proportion-/+ margin of error =0.459 ; 0.501

hence 45.9% to 50.1%

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