Question

An online site presented this​ question, "Would the recent norovirus outbreak deter you from taking a​ cruise?" Among the

34 comma 303

people who​ responded,

69

​%

answered​ "yes." Use the sample data to construct a

90

​%

confidence interval estimate for the proportion of the population of all people who would respond​ "yes" to that question. Does the confidence interval provide a good estimate of the population​ proportion?

Answer 1

z_α/2 = 1.6448536251

Lower Bound = p̂ - z_α/2•√p̂(1 - p̂)/n = 0.69 - (1.6448536251)(0.0024971195836803327) = 0.6858926038

Upper Bound = p̂ + z_α/2•√p̂(1 - p̂)/n = 0.69 + (1.6448536251)(0.0024971195836803327) = 0.6941073962

Confidence Interval = (0.686, 0.694)

Does the confidence interval provide a good estimate of the population​ proportion?

YES

For a population with unknown population proportion (p):

a) Form all possible samples of size 34303 and calculate sample proportion (p̂) of each sample.

b) For each sample proportion, calculate a confidence interval; hence, there are many, many confidence intervals.

c) Approximately 90% of these confidence intervals will contain the population proportion (p).

d) Since we do not know if the confidence interval (0.6858926038, 0.6941073962) contains the
population proportion (p) or not, we are only 90% confident that (0.6858926038, 0.6941073962) contains
the population proportion (p).