"\\text{The confidence interval for the }\\\\\n\\text{proportion can be calculated as follows:}\\\\\n \\hat p \u00b1Z_{\\frac{\u03b1}{2}}(\\sqrt {\\frac{\\hat p (1-\\hat p )}{n}}),\\\\\n \\hat p =\\frac{20}{25}=0.8,\\\\\n Z_{\\frac{\u03b1}{2}}=Z_{\\frac{0.05}{2}}=Z_{0.025}=1.96,\\\\\n \\text{the lower limit }= 0.8-1.96(\\sqrt {\\frac{0.8 (1-0.8 )}{25}})\\\\\n\u22480.6432,\\\\\n \\text{the upper limit }= 0.8+1.96(\\sqrt {\\frac{0.8 (1-0.8 )}{25}})\\\\\n\u22480.9568,\\\\\n \\text{so the confidence interval is}\\\\\n 0.6432\\leq \\ p \\leq 0.9568"
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