Example: As 483 of 3333 customers churned, the estimate of the population proportion of all the customers that churned is: =(number who churn)/(sample size)=/=483/3333=0.1449 Testing at =0.10 whether differs from 0.15 (0.14949) , the hypotheses are: _0: =0.15 . _: 0.15 The test statistics is: _= (_0)/((_0 (1_0))/)=(0.14490.15)/((0.15(0.85))/3333)=0.8246 As _<0, the p\value=2(<_) = 2(<0.8246)=20.2048=0.4096 Since the p-value is not less than =0.10 we would not reject H0 Assessing the strength of evidence using Table 4.5 we find no evidence against H0. Also, given a confidence interval, we may perform two-tailed hypothesis test for as we did for