Question

web based software company is interested in estimating the proportion of individuals who use the Firefox browser. In a sample of 200 of individuals, 30 users stated that they used Firefox. Using this data, construct a 99% confidence interval for the proportion of all individuals that use Firefox. What is the lower limit on the 99% confidence interval? Give your answer to three decimal places.

Answer 1

Answer with explanation:

Let $\hat{p}$ denotes the sample proportion.

As per given , we have

n= 200

$\hat{p}=\dfrac{30}{200}=0.15$

Critical value for 99% confidence : $z_{\alpha/2}=2.576$

Confidence interval for population proportion :-

$\hat{p}\pm z_{\alpha/2} \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.15\pm (2.576)(\sqrt{\dfrac{0.15(1-0.15)}{200}})\\\\\approx0.15\pm0.065\\\\=(0.15-0.065,\ 0.15+0.065)\\\\=(0.085,\ 0.215)$

Hence, a 99% confidence interval for the proportion of all individuals that use Firefox: $(0.085,\ 0.215)$

The lower limit on the 99% confidence interval = 0.085