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

Answer:

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

(0.3342 , 0.5258)

Step-by-step 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

                           $p^{-} = \frac{x}{n} = \frac{43}{100} = 0.43$

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

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

$(0.43 -1.96 \sqrt{\frac{0.43(1-0.43)}{100} } ,0.43 +1.96 \sqrt{\frac{0.43(1-0.43)}{100} })$

( 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)