Answer to Question #199152 in Statistics and Probability for mandona

Question

EMC research conducted a poll between Jan. 14 and Jan 22 2014. They asked a SRS of 805 voting Seattlites if they supported the minimum raise hike to $15/hr. They found a sample proportion of 68%. What is a 95% confidence interval for the true proportion? 2. EMC research conducted a poll between Jan. 14 and Jan 22 2014. They asked a SRS of 805 voting Seattlites if they supported the minimum raise hike to $15/hr. They found a sample proportion of 68%. What is a 90% confidence interval for the true proportion?

Answer 1

1.

95% confidence interval:

$p_0-Z_{0.95}\sqrt{\frac{p_0(1-p_0)}{n}}<p<p_0+Z_{0.95}\sqrt{\frac{p_0(1-p_0)}{n}}$

where $p_0=0.68,n=805$

$p_0-Z_{0.95}\sqrt{\frac{p_0(1-p_0)}{n}}=0.68-1.96\sqrt{\frac{0.68\cdot0.32}{805}}=0.65$

$p_0+Z_{0.95}\sqrt{\frac{p_0(1-p_0)}{n}}=0.68+1.96\sqrt{\frac{0.68\cdot0.32}{805}}=0.71$

$0.65<p<0.71$

2.

90% confidence interval:

$p_0-Z_{0.9}\sqrt{\frac{p_0(1-p_0)}{n}}<p<p_0+Z_{0.9}\sqrt{\frac{p_0(1-p_0)}{n}}$

where $p_0=0.68,n=805$

$p_0-Z_{0.9}\sqrt{\frac{p_0(1-p_0)}{n}}=0.68-1.645\sqrt{\frac{0.68\cdot0.32}{805}}=0.653$

$p_0+Z_{0.9}\sqrt{\frac{p_0(1-p_0)}{n}}=0.68+1.645\sqrt{\frac{0.68\cdot0.32}{805}}=0.707$

$0.653<p<0.707$