Question

In an episode of the old school version of the game show Family Feud, 43 out of a random sample of 100 people said they pick their noses at red lights. Find a 95% confidence interval of the proportion of all people who pick their noses at red lights.

Answer 1

95% of confidence interval of the proportion of all people who pick their noses at red lights

(0.3342 , 0.5258)

Explanation:

Step(i):-

Given sample size 'n' = 100

Given data 43 out of a random sample of 100 people said they pick their noses at red lights.

sample proportion

`p^(-) = (x)/(n) = (43)/(100) = 0.43`

Level of significance = 0.05

Z₀.₀₅ = 1.96

Step(ii):-

95% of confidence interval of the proportion of all people who pick their noses at red lights

`(p^(-) -Z_(\alpha ) \sqrt{(p(1-p))/(n) } ,p^(-) +Z_(\alpha ) \sqrt{(p(1-p))/(n) })`

`(0.43 -1.96 \sqrt{(0.43(1-0.43))/(100) } ,0.43 +1.96 \sqrt{(0.43(1-0.43))/(100) })`

( 0.43 - 0.0958 , 0.43 + 0.0958)

(0.3342 , 0.5258)

Conclusion:-

95% of confidence interval of the proportion of all people who pick their noses at red lights

(0.3342 , 0.5258)