Answer:
The 95% confidence interval for the proportion opposing health care changes is (0.6082, 0.6918).
Step-by-step explanation:
The (1 - α)% confidence interval for the population proportion is:
[tex]CI=\hat p\pm z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
The information provided is:
[tex]\hat p=0.65\\n=500\\\text{Confidence level}=95\%[/tex]
The critical value of z for 95% confidence level is:
[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]
*Use a z-table.
Compute the 95% confidence interval for the proportion opposing health care changes as follows:
[tex]CI=\hat p\pm z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
[tex]=0.65\pm 1.96\sqrt{\frac{0.65(1-0.65)}{500}}\\\\=0.65\pm 0.04181\\\\=(0.60819, 0.69181)\\\\\approx (0.6082, 0.6918)[/tex]
Thus, the 95% confidence interval for the proportion opposing health care changes is (0.6082, 0.6918).
