Respuesta :
Answer:
90% confidence interval for the proportion of fans who bought food from the concession stand
(0.5603,0.6529)
Step-by-step explanation:
Step(i):-
Given sample size 'n' =300
Given data random sample of 300 attendees of a minor league baseball game, 182 said that they bought food from the concession stand.
Given sample proportion
[tex]p^{-} = \frac{x}{n} = \frac{182}{300} =0.606[/tex]
level of significance = 90% or 0.10
Z₀.₁₀ = 1.645
90% confidence interval for the proportion is determined by
[tex](p^{-} - Z_{0.10}\sqrt{\frac{p(1-p)}{n} } , p^{-} +Z_{0.10}\sqrt{\frac{p(1-p)}{n} })[/tex]
[tex](0.6066 - 1.645\sqrt{\frac{0.6066(1-0.6066)}{300} } ,0.6066+1.645\sqrt{\frac{0.6066(1-0.6066)}{300} })[/tex]
(0.6066 - 0.0463 ,0.6066 + 0.0463)
(0.5603,0.6529)
final answer:-
90% confidence interval for the proportion of fans who bought food from the concession stand
(0.5603,0.6529)
