Question

(1 point) 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. The proportion of all BlueSky Air flights that arrive on time. B. All BlueSky Air flights. C. The proportion of the 160 BlueSky Air flights that arrive on time. D. Whether a BlueSky Air flight arrives on time. Part ii) Let p p be the population proportion of flights that arrive on time. What are the null and alternative hypotheses? A. Null: p=0.80 p=0.80 . Alternative: p<0.80 p<0.80 . B. Null: p>0.80 p>0.80 . Alternative: p≤0.80 p≤0.80 . C. Null: p=0.80 p=0.80 . Alternative: p≠0.80 p≠0.80 . D. Null: p=0.80 p=0.80 . Alternative: p>0.80 p>0.80 . E. Null: p=115/160 p=115/160 . Alternative: p≠115/160 p≠115/160 . F. Null: p<0.80 p<0.80 . Alternative: p≥0.80 p≥0.80 . Part iii) Compute the P-value (please round to four decimal places): Part iv) The P 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. 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. C. The observed proportion of flights that arrive on time is unusually high if BlueSky Air's claim is false. D. There is a [P-value] probability that BlueSky Air's claim is true. E. The observed proportion of flights that arrive on time is unusually low if BlueSky Air's claim is false. F. The observed proportion of flights that arrive on time is unusually high 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 true. Part v) What is an appropriate conclusion for the hypothesis test at the 1% significance level? A. There is sufficient evidence to contradict BlueSky Air's claim. B. BlueSky Air's claim is true. C. BlueSky Air's claim is false. D. There is sufficient evidence to support BlueSky Air's claim. Part vi) Which of the following scenarios describe the Type I error of the test? A. The data suggest that BlueSky Air's claim is false when in fact the claim is true. B. The data suggest that BlueSky Air's claim is true when in fact the claim is true. C. The data suggest that BlueSky Air's claim is false when in fact the claim is false. D. The data suggest that BlueSky Air's claim is true when in fact the claim is false. Part vii) Based on the result of the hypothesis test, which of the following types of errors are we in a position of committing? A. Both Type I and Type II errors. B. Type I error only. C. Neither Type I nor Type II errors. D. Type II error only.

Answer #1

i)The proportion of all BlueSky Air flights that arrive on time.

ii)

Null: p=0.80

Alternative: p<0.80

iii)

pvalue =0.0051 ( please try 0.0052 if this comes wrong and revert)

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.

v)

A. There is sufficient evidence to contradict BlueSky Air's claim

vi)

A. The data suggest that BlueSky Air's claim is false when in fact the claim is true

B. Type I error only.

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
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below what the airline claims)?
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