Question

In a recent​ poll, 784 adults were asked to identify their favorite seat when they​ fly, and 517 of them chose a window seat. Use a 0.01 significance level to test the claim that the majority of adults prefer window seats when they fly. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method and the normal distribution as an approximation to the binomial distribution.

1. What is the test statistic? z=______

2. What is the P value? p=_____

Answer 1

Ho: majority of adults dont prefer window seats when they fly. p = 50%
h1: majority of adults prefer window seats when they fly. p> 50%

X=   517  
n=   784  
p-hat = X/n=   0.659    =517/784

test statistic, z = (phat-p)/sqrt(p*(1-p)/n)
z = (0.659-0.5)/SQRT(0.5*(1-0.5)/784)
z = 8.904

p-value
1-P(Z<z)
1-P(z<8.904)
=1-NORMSDIST(8.904)
0.00000

With z = 8.904, p<5%, I reject ho and conclude that the majority of adults prefer window seats when they fly. p> 50%.