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Free_Kalibri [48]
2 years ago
8

A randomized study compared two different systems for tracking baggage at an airport. The treatment group using System A reporte

d a mean of 18.5 lost bags per day. The treatment group using System B reported a mean of 16.6 lost bags per day.
To determine if the results are significant, the data are re-randomized and the differences of the means are shown in the dot plot.


What is the best conclusion to make based on the data?

a) The difference of the means is not significant because the re-randomizations show that it is outside the range of what could happen by chance.

b) The difference of the means is significant because the re-randomizations show that it is within the range of what could happen by chance.

c) The difference of the means is significant because the re-randomizations show that it is outside the range of what could happen by chance.

d) The difference of the means is not significant because the re-randomizations show that it is within the range of what could happen by chance.

Mathematics
2 answers:
Natasha2012 [34]2 years ago
8 0

Answer:

D. The difference of the means is not significant because the re-randomizations show that it is within the range of what could happen by chance.

Step-by-step explanation:

The treatment group using System A reported a mean of 18.5 lost bags per day. The treatment group using System B reported a mean of 16.6 lost bags per day.

The best conclusion that can be made is - The difference of the means is not significant because the re-randomizations show that it is within the range of what could happen by chance.

As we know, in statistics, nothing happens by chance. So, this option is correct.

geniusboy [140]2 years ago
3 0

Answer:

It is D

Step-by-step explanation:

took the test

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Row total for no phone call = 8 +6 = 14

To calculate their respective row relative frequencies, let's use:

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Now, the two-way frequency table will be computed as:

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Desirable behavior = \frac{19}{28} = 0.67857 ≈0.69

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Desirable behaviour = \frac{8}{14} = 0.5714 ≈ 0.57

Undesirable behaviour = \frac{6}{14} = 0.4286 ≈ 0.43

The complete two-way table is attached.

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