Question

Janet wants to estimate the percentage of people who have been required to take a drug test for work. She surveys 250 individuals and finds that 175 have taken a drug test for work. Identify the values needed to calculate a confidence interval at the 98% confidence level. Then find the confidence interval.

Answer 1

Answer:

98% of  confidence interval for the true population

(0.6326 , 0.7674)

Step-by-step explanation:

Step(i):-

Given sample size 'n' = 250

She surveys 250 individuals and finds that 175 have taken a drug test for work.

sample proportion

                $p = \frac{x}{n} = \frac{175}{250} =0.7$

level of significance ∝ = 0.05

98% of  confidence interval for the true population is determined by

$(p^{-} - Z_{0.02} \sqrt{\frac{p(1-p)}{n} } ,p +Z_{0.02} \sqrt{\frac{p(1-p)}{n} } )$

Z₀.₀₂ = 2.326

Step(ii):-

98% of  confidence interval for true population is determined by

$(0.7 - 2.326\sqrt{\frac{0.7(1-0.7)}{250} } ,0.7 + 2.326 \sqrt{\frac{0.7(1-0.7)}{250} } )$

( 0.7 -  0.0674 , 0.7 + 0.0674)

(0.6326 , 0.7674)

Conclusion:-

98% of  confidence interval for the true population is determined by

(0.6326 , 0.7674)