Question

A study involves 669 randomly selected deaths, with 31 of them caused by accidents. construct a 98% confidence interval for the true percentage of all deaths that are caused by accidents.

Answer 1

Answer:

Confidence Interval : 2.72% < p < 6.48%

Step-by-step explanation:

Number of randomly selected deaths = 669

Number of deaths caused by accident = 31

$\text{Sample Proportion, p = }\frac{31}{669}=0.046$

z-score for 98% confidence level = 2.326

$\text{Margin of Error = }z\times \sqrt{\frac{p\cdot (1-p)}{n}}\\\\\text{Margin of error = }2.326\times \sqrt{\frac{0.046\times 0.954}{669}}\\\\\implies\text{Margin of error = }0.0188$

Confidence Interval : 0.046 - 0.0188 < p < 0.046 + 0.0188

                                 : 0.0272 < p < 0.0648

                                 : 2.72% < p < 6.48%

Hence, Confidence Interval : 2.72% < p < 6.48%