document.write( "Question 1206720: Use the empirical rule to solve the following problems below Assume all the data follow the normal distribution: A researcher is studying the proportion of adults who support a particular policy. A random sample of 200 adults was selected, and 150 of them indicated support for the policy. a. Find a 95% confidence interval for the true proportion of adults who support the policy.
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Algebra.Com's Answer #844329 by MathLover1(20850)\"\" \"About 
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if a random sample of \"200\" adults was selected, and \"150+\"of them indicated support for the policy, we have\r
\n" ); document.write( "\n" ); document.write( "\"p=150%2F200=3%2F4=0.75\"\r
\n" ); document.write( "\n" ); document.write( " a \"95\"% confidence=> \"z=1.96\"\r
\n" ); document.write( "\n" ); document.write( "then, confidence interval for the true proportion of adults who support the policy is\r
\n" ); document.write( "\n" ); document.write( "\"p%2B-z%2Asqrt%28p%281-p%29%2Fn%29\"\r
\n" ); document.write( "\n" ); document.write( "\"0.75%2B-1.96%2Asqrt%280.75%281-0.75%29%2F200%29\"\r
\n" ); document.write( "\n" ); document.write( "\"0.75%2B-0.06001249869818786\"\r
\n" ); document.write( "\n" ); document.write( "\"0.75%2B-0.06\"\r
\n" ); document.write( "\n" ); document.write( "\"0.75-0.06=0.69\"\r
\n" ); document.write( "\n" ); document.write( "\"0.75%2B0.06=0.81\"\r
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\n" ); document.write( "\n" ); document.write( "we are \"95\"% confident there will be between \"69\"% and \"81\"% adults, with a margin of error of \"0.06\", who support the policy\r
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