Question

Construct a 90% confidence interval of the population proportion using the given information X =120, n= 300The lower bound isThe upper bound is

Answer 1

Solution

For this case we have the following data:

x= 120 , n= 300

The sample proportion is given by:

$p_x=\frac{X}{n}=\frac{120}{300}=\frac{2}{5}^{}$

The confidence interval for a proportion is given by:

$P_x\pm z_{\frac{\alpha}{2}}\cdot\sqrt[]{\frac{P_x(1-P_x)}{n}}$

The critical value for 90% confidence given is z= 1.645 and replacing we have:

$\frac{2}{5}\pm1.645\cdot\sqrt[]{\frac{\frac{2}{5}(1-\frac{2}{5})}{300}}$

Then the confidence interval is:

(0.353; 0.447)