Question

I believe that the number of adults who eat beef on the 4th of July in Texas is greater than the number of adults who eat beef on the 4th of July in California. In two independent polls, I find that 420 out of 500 Texans ate beef on the 4th of July, and 380 out of 500 Californians ate beef on the 4th of July. Please test with a 95% confidence interval.

Can anyone help me solve this step by step, please? I don't want just an answer. Thank you very much.

Answer 1

p̂1 = 420 / 500 = 0.84
p̂2 = 380 / 500 = 0.76

(p̂1 - p̂2) ± Z(α/2) * √( ((p̂1 * q̂1)/ n1) + ((p̂2 * q̂2)/ n2) )
Z(α/2) = Z(0.05 /2) = 1.96
Lower Limit = ( 0.84 - 0.76 )- Z(0.05/2) * √(((0.84 * 0.16 )/ 500 ) + ((0.76 * 0.24 )/ 500 ) = 0.0307
upper Limit = ( 0.84 - 0.76 )+ Z(0.05/2) * √(((0.84 * 0.16 )/ 500 ) + ((0.76 * 0.24 )/ 500 )) = 0.1293
95% Confidence interval is ( 0.0307 , 0.1293 )
( 0.0307 < ( P1 - P2 ) < 0.1293 )

Since the interval contain only positive values, we can conclude that the number of adults who eat beef on the 4th of July in Texas is greater than the number of adults who eat beef on the 4th of July in California.