Question

In a recent​ poll,

805805

adults were asked to identify their favorite seat when they​ fly, and

475

of them chose a window seat. Use a

0.05

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.

Answer 1

Solution:

Here, we have to use z test for population proportion.

Null hypothesis: H₀: No majority of adults prefer window seats when they fly.

Alternative hypothesis: H_a: Majority of adults prefer window seats when they fly.

H₀: p ≤ 0.5 versus H_a: p > 0.5

This is an upper tailed test.

We are given

Level of significance = α = 0.05

X = 475

n = 805

Sample proportion = P = X/n = 475/805 = 0.59

Test statistic formula is given as below:

Z = (P – p)/sqrt(pq/n)

Where, q = 1 – p = 1 – 0.5 = 0.5

Z = (0.59 - 0.50)/sqrt(0.5*0.5/805)

Z = 5.107054

Z = 5.11

P-value = 0.0000

(by using z-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that Majority of adults prefer window seats when they fly.