Question

A travel journal claims that an airline's on-time arrival rate is only 78%, of less. The airline wants to prove the journal wrong. Airline takes a random sample of 750 flights and finds that 615 were on time. Test at the 4% level of significance if there is sufficient evidence to prove the journal wrong. State the null and alternative hypotheses and report the conclusion.

a. calculate the p-value

Answer 1

Solution :

Given that,

$p_{0}$ = 0.78

1 - $p_{0}$ = 0.22

n = 750

x = 615

Level of significance = $\alpha$ = 0.04

Point estimate = sample proportion = $\hat p$ = x / n = 0.82

This a two- tailed test.

The null and alternative hypothesis is,

Ho: p = 0.82

Ha: p $\neq$ 0.82

Test statistics

z = ($\hat p$ - $p_{0}$ ) / $\sqrt{}$ $p_{0}$*(1-$p_{0}$) / n

= ( 0.82 - 0.78) / $\sqrt{}$ (0.78*0.22) /750

= 2.644

a)

P-value = 2 * P(Z > z )

= 2 * ( 1 - P(Z < 2.644 ))

= 2 * 0.0041

= 0.0082

The p-value is p = 0.0082, and since p = 0.0082 < 0.04, it is concluded that the null hypothesis is rejected.