Consider a flight to be on time if it arrives no later than 15 minutes after the scheduled arrival time. The data below represents the arrival delay time in minutes. Negative numbers correspond to flights that arrived early (before the scheduled time). Use a significant level of 0.05 and the data to test the claim made by CNN that 79.5% of flights are on time. -32 -25 -26 -6 5 -15 -17 -36 -29 -18 -12 -35 2 -33 -5 0 0 -1 -33 -5 -14 -39 -21 -32 -5 -32 -13 -9 -19 49 -30 -23 14 -21 -32 11 -23 28 103 -19 -5 -46 13 -3 13 106 -34 -24

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AlonsoDehner AlonsoDehner · Answer 1 · 2019-11-26T08:09:20+01:00

Answer:

Claim is true at 5% level

Step-by-step explanation:

Given is the data about the arrival of flights. Negative numbers correspond to flights that arrived early (before the scheduled time).

Let p be the proportion of flights on time

[tex]H_0: p = 0.795\\H_a: p \neq 0.795[/tex]

(Two tailed test at 5% significant level)

Since upto 15 minutes is considered on time, from the data of 48 flights we find that number of flights arrive delay is 4

Proportion of sample for flights arrived on time = [tex]\frac{44}{48} =0.917[/tex]

Assuming null hypothesis to be true, std error = [tex]\sqrt{\frac{0.795(1-0.795)}{\sqrt{48} } } \\=0.0583[/tex]

Z statistic = p diff/se =2.093

Since this value lies between -1.96 and 1.96 we accept H0

The claim that 79.5% of flights are on time is proved at 5% level.