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)?

Part i) What is the parameter of interest? A. All BlueSky Air flights. B. The proportion of all BlueSky Air flights that arrive on time. C. Whether a BlueSky Air flight arrives on time. D. The proportion of the 160 BlueSky Air flights that arrive on time.

Part ii) Let pp be the population proportion of flights that arrive on time. What are the null and alternative hypotheses? A. Null: p=0.80p=0.80. Alternative: p0.80p0.80. B. Null: p=0.80p=0.80. Alternative: p>0.80p>0.80. C. Null: p=0.80p=0.80. Alternative: p<0.80p<0.80. D. Null: p=115/160p=115/160. Alternative: p115/160p115/160. E. Null: p>0.80p>0.80. Alternative: p0.80p0.80. F. Null: p<0.80p<0.80. Alternative: p0.80p0.80.

Part iii) Compute the z-statistic:

Part iv) Compute the P-value:

Part v) What is an appropriate conclusion for the hypothesis test at the 1% significance level? A. BlueSky Air's claim is true. B. BlueSky Air's claim is false. C. This is not significant evidence to contradict BlueSky Air's claim. D. This is significant evidence to contradict BlueSky Air's claim.