Question

Candy has observed 250 sports cars at an intersection and found that 15% did not make a complete stop. Candy tells you that the margin of error was 8%. What was the confidence level of the study?

Answer 1

Answer:

The confidence level of the study is 99.9%

The confidence level is 99.9% of the margin error is 8%

Step-by-step explanation:

Explanation:-

Candy has observed 250 sports cars

The sample size 'n' = 250

Sample proportion 'p' = 15% = 0.15

The margin of error = 8% = 0.08

The margin of error  is determined by   = $\frac{z_{\alpha }\sqrt{p(1-p)} }{\sqrt{n} }$

                                                 $0.08 X \sqrt{250} = z_{\alpha } \sqrt{0.15 (1-0.15)}$

                                                    $\frac{ 0.08 X \sqrt{250}}{ \sqrt{0.15 (1-0.15)}} = z_{\alpha }$

                                                    $z_{\alpha }= \frac{ 0.08 X \sqrt{250}}{ \sqrt{0.15 (1-0.15)}} = \frac{1.2649}{0.35707}$

                                                    $z_{\alpha } = 3.5431$

Conclusion:-

The z- score of 99.9% level of significance = 3.54 for two tailed test

Hence the confidence level is 99.9% of the margin error is 8%.