Question

In a random sample of 300 attendees of a minor league baseball game, 182 said that they bought food from the concession stand. Create a 90% confidence interval for the proportion of fans who bought food from the concession stand. Use a TI-83, TI-83 plus, or TI-84 calculator, rounding your answers to three decimal places.

Answer 1

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

                       $p^{-} = \frac{x}{n} = \frac{182}{300} =0.606$

level of significance = 90% or 0.10

Z₀.₁₀ = 1.645

90% confidence interval for the proportion is determined by

$(p^{-} - Z_{0.10}\sqrt{\frac{p(1-p)}{n} } , p^{-} +Z_{0.10}\sqrt{\frac{p(1-p)}{n} })$

$(0.6066 - 1.645\sqrt{\frac{0.6066(1-0.6066)}{300} } ,0.6066+1.645\sqrt{\frac{0.6066(1-0.6066)}{300} })$

(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)