Question

Harris Interactive® conducted a poll of American adults in August of 2011 to study the use of online medical information. Of the 1,019 randomly chosen adults, 60% had used the Internet within the past month to obtain medical information. Use the results of this survey to create an approximate 95% confidence interval estimate for the percentage of all American adults who have used the Internet to obtain medical information in the past month.

Answer 1

Answer:

$0.60 - 1.96\sqrt{\frac{0.60(1-0.60)}{1019}}=0.570$

$0.60 + 1.96\sqrt{\frac{0.60(1-0.60)}{1019}}=0.630$

The 95% confidence interval for the true proportion would be given by (0.570;0.630) .

And if we convert this into % we got (57.0%, 63.0%)

Step-by-step explanation:

The information given we have the following info given:

$n = 1019$ represent the sampel size

$\hat p=0.6$ represent the sample proportion of interest

The confidence level is 95%, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

Replacing the info given we got:

$0.60 - 1.96\sqrt{\frac{0.60(1-0.60)}{1019}}=0.570$

$0.60 + 1.96\sqrt{\frac{0.60(1-0.60)}{1019}}=0.630$

The 95% confidence interval for the true proportion would be given by (0.570;0.630) .

And if we convert this into % we got (57.0%, 63.0%)