SOLUTION: Could someone please explain this to me? I'm having a really hard time understanding how to work them out? 1. In a sample of 200 people, 76 people would rather work out at hom

Question 1078523: Could someone please explain this to me? I'm having a really hard time understanding how to work them out?
1. In a sample of 200 people, 76 people would rather work out at home than in a gym. Find the 99% confidence interval for the true proportion of people who would rather work out at home than in a gym for the entire population.
2. A study found that out of 300 people 60% of them prefer to eat hamburgers rather than hot dogs. Fin the 95% confidence interval for the true proportion of people who prefer to eat hamburgers rather than hot dogs in the entire population.

Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
1) The sample size is 200 which is sufficient(greater than 30) to use a normal distribution
:
The sample statistic proportion is (76 / 200) = 0.38
:
The standard error(SE) for the sample = square root(0.38 * (1-0.38) / 200) = 0.0343
:
alpha (a): a = 1 - (confidence level / 100) = 1 - (99/100) = 0.01
p* = 1 - a/2 = 1 - 0.01/2 = 0.995
critical value(CV) is determined by looking for the z-value associated with p*
CV = 2.57
margin of error(ME) = CV * SE = 2.57 * 0.0343 = 0.0882 approximately 0.09
confidence interval(CI) = sample proportion + or - margin of error
CI = 0.38 + or - 0.09 = (0.29, 0.47)
:
2) The sample size is 300 which is sufficient(greater than 30) to use a normal distribution
:
The sample statistic proportion = 0.60
:
SE = square root(0.60 * (1-0.60) / 300) = 0.0283
:
alpha (a): a = 1 - (confidence level / 100) = 1 - (95/100) = 0.05
p* = 1 - a/2 = 1 - 0.05/2 = 0.975
critical value(CV) is determined by looking for the z-value associated with p*
CV = 1.96
ME = CV * SE = 1.96 * 0.0283 = 0.0555 approximately 0.06
CI = 0.60 + or - 0.06 = (0.54, 0.66)
:

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