"Question\\; 7\\\\\nLet\\;\\hat p=\\frac{34}{200}=0.17\\\\\nThen,\\; \\text{the confidence interval for the proportion }\\\\\n\\text{can be calculated as follows:}\\\\\n \\hat p\u00b1Z_{\\frac{\u03b1}{2}}(\\sqrt{\\frac{\\hat p(1-\\hat p)}{n}}),\\\\\n\\hat p=0.17\\\\\n Z_{\\frac{\u03b1}{2}}=Z_{\\frac{0.05}{2}}=Z_{0.025}=1.96,\\\\\n n=200\\\\\n \\text{ the lower limit }= 0.17-1.96(\\sqrt{\\frac{0.17(1-0.17)}{200}})\\\\\n=0.1179\\\\\n \\text{ the upper limit} = 0.17+1.96(\\sqrt{\\frac{0.17(1-0.17)}{200}})\\\\\n=0.2221\\\\\n\\text{the answer is (5) none of the above}\\\\\nQuestion \\;8\\\\\n Z_{\\frac{\u03b1}{2}}=Z_{\\frac{0.01}{2}}=Z_{0.005}=2.58,\\\\\n\n \\text{ the upper limit} = 0.17+2.58(\\sqrt{\\frac{0.17(1-0.17)}{200}})\\\\\n=0.2385\\\\\n\\text{the answer is (4) 0.2385}\\\\"
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