A few days ago
Help with Statistics?
1025 randomly selected adults were surveyed and 29% of them said that they eat McDonalds at least two times a year. Assume that all conditions for a one-proportion z-interval are met and answer the following:
Create a 95% confidence interval for the % of people that will eat at McDonalds a minimum of twice a year.
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A few days ago
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a 95% CI for a proportion p is:
pHat +- 1.96*Sqrt(pHat(1-pHat)/(n))
now, this is a standard method, there is a better method but the sample size here will make it irrelevant. the 1.96 is the standard score from the standard normal distribution such that 95% of the data is between -1.96 and +1.96.
0.29 +- 1.96*Sqrt(0.29*(1-0.29)/1025) =
{0.26222, 0.31778}
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