Of 98 adults selected randomly from one town, 64 have health insurance. Find a 9
ID: 3223285 • Letter: O
Question
Of 98 adults selected randomly from one town, 64 have health insurance. Find a 90% confidence interval for the proportion of all adults in the town who have health insurance. Assume that you wish to estimate a population proportion, p. For the given margin of error and confidence level, determine the sample size required. You wish to estimate the proportion of adults that have ever used alternative medicine. Obtain a sample size that will ensure a margin of error of at most 0.01 for a 95% confidence interval. It deemed reasonable to presume that of those sampled, the proportion that have used alternative medicine will be at least 0.64.Explanation / Answer
12. We will follow below steps to calculate confidence intervals:
SE = sqrt [ p(1 - p) / n ] = sqrt [ (0.35)*(0.65) / 98 ] = sqrt [ 0.0023 ] = 0.048
Therefore, the 90% confidence interval is the range defined by 0.65 + 0.08.
13. Margin of error= 0.01
SE= sqrt [ p(1 - p) / n ]= sqrt [ 0.64(1 - 0.64) / n ]=sqrt [ 0.23 / n ]
critical value *SE =0.01
critical value*sqrt [ 0.23 / n ]=0.01
(t value for given degrees of freedom)*sqrt [ 1 / n ]=0.02
Through hit and trial, we will find that for a sample size of n~ 6250, we are getting the desired results for given confidence interval.
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