8.125 An alternative interval for the population proportion The large-sample confidence interval for a proportion substitutes pn for the unknown value of p in the formula for the standard deviation of pn. An alternative method to form a 95% confidence interval determines the endpoints of the interval by finding the values for p that are 1.96 standard deviations from the sample proportion, without a need to estimate the standard deviation. To do this, you solve for p in the equation

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|pp|=1.96p(1 - p)/n. This interval is known as the (Wilson) score confidence interval for a proportion. a. For Example 10 with no students without an MP3 player in a sample of size 20, substitute p 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 1.96 (se). b. Which confidence interval seems more believable? Why?