Hi Ho: p ≥ .65 claim Ha: p < .65 z(140/200) = ( .7 - .65)/sqrt[.65*.35/200] = 1.4825 p-value = P(z > 1.4825) = normalcdf(1.4825,100) = 0.069 Conclusion:: Since the p-value is greater than 1%, fail to reject Ho. there is sufficient evidence at the 1% significance level to indicate that the population proportion of successes is at least 65% Wish You the Best in your Studies.