The large sample confidence interval for a proportion substitutes p for
The large-sample confidence interval for a proportion substitutes p̂ for the unknown value of p in the exact standard error of p̂. A less approximate 95% confidence interval has endpoints determined by the p values that are 1.96 standard errors from the sample proportion, without estimating the standard error. To do this, you solve for p in the equation
| p̂ - p | = 1.961√p(1 - p)/n.
a. For Example 11 with no students without iPods in a sample of size 20, substitute pn and n in this equation and show that the equation is satisfied at p = 0.83337 and at p = 1. So the confidence interval is (0.83887, 1), compared to (1, 1) with p̂ ± 1.96(se).
b. Which confidence interval seems more believable? Why?
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