Question

BlueSky Air claims that at least 80% of its flights arrive on time. A random sample of 160 BlueSky Air flights revealed that 115 arrive on time. Do the data provide sufficient evidence to contradict the claim by BlueSky Air (i.e., you would like to see whether the percentage of the airline's flight is below what the airline claims)?

The P-value is computed to be [P-value] (your answer in Part iii). Using all of the information available to you, which of the following is/are correct? (check all that apply)

A. Assuming BlueSky's claim is true, there is a [P-value] probability that in a random sample of 160 flights, 115 or fewer flights arrive on time.
B. The observed proportion of flights that arrive on time is unusually high if BlueSky Air's claim is false.
C. There is a [P-value] probability that BlueSky Air's claim is true.
D. Assuming BlueSky's claim is false, there is a [P-value] probability that in a random sample of 160 flights, 115 or fewer flights arrive on time.
E. The observed proportion of flights that arrive on time is unusually high if BlueSky Air's claim is true.
F. The observed proportion of flights that arrive on time is unusually low if BlueSky Air's claim is true.
G. The observed proportion of flights that arrive on time is unusually low if BlueSky Air's claim is false.

Answer 1

H0:Null Hypothesis: P 0.80 (Claim)

HA: Alternative Hypothesis: P < 0.80

n = Sample Size = 160

P= Population Proportion = 0.80

Q = 1 - P = 0.20

SE =

p = Sample Proportion = 115/160 = 0.7188

Test statistic s given by:
Z = (0.7188 - 0.80)/0.0316

= - 2.5712

Table of Area Under Standard Normal Curve gives area = 0.4949

P - Value = 0.5 - 0.4949 = 0.0051

Correct option:

A. Assuming BlueSky's claim is true, there is a 0.0051 probability that in a random sample of 150 BlueSky Air flights, 115 or fewer flights arrive on time.