Question

EXPLANATION NEEDED

1. A travel agency claims that at least 58% of all airline bookings are made online. However a sample of 70 airline reservations revealed that only 52% of recent bookings were made online. At the 10% significance level, carry out the hypothesis test.

A. First, highlight the null and alternative hypotheses.

H₀: p ≤ 0.58 or 58%; H₁: p > 0.58 or 58%

H₀: p ≥ 0.58 or 58%; H₁: p < 0.58 or 58%

B. Describe the normal curve appropriate for this test of hypothesis

Two-tail with critical Z values of ±1.285
       One-tail (right-tail) with critical Z value of 1.285
       Two-tail with critical Z values of ±1.285
       One-tail (left-tail) with critical Z value of -1.285
       None of the above

Z = -1.0171

P = about 0.15

C. State your decision and conclusion
       Do not reject H0. There is evidence the claim is wrong.
       Reject H0. There is no evidence to reject the claim.
       Do not reject H0. There is no evidence to reject the claim.
       None of the above

Answer 1

Solution :

This is the left tailed test .

The null and alternative hypothesis is

H₀ : p 0.58

H_a : p > 0.58

n = 70

= 0.52

P₀ = 0.58

1 - P₀ = 1 - 0.58 = 0.42

z = - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.52 - 0.58 / [(0.58 * 0.42) /70 ]

= -1.0171

Test statistic = -1.0171

= 0.10

Z= Z_0.10 = -1.285

P(z < -1.0171) = 0.15

P-value = 0.15

= 0.10

P-value >

Do not reject the null hypothesis .

Do not reject H0. There is no evidence to reject the claim.