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Carter wants to know if warming up will help runners sprint faster. Thirty track and field athletes volunteered to participate in his study. He randomly assigns 15 athletes to warm-up for 10 minutes. All 30 participants sprint the same distance. He calculates the mean for each group and determines that the mean for the warm-up group was 10.7 seconds, and the mean for the other group was 13.2 seconds. To test the difference of means he re-randomized the data 54 times, and the differences are plotted in the dot plot below. What can Carter conclude from his study?

Carter wants to know if warming up will help runners sprint faster Thirty track and field athletes volunteered to participate in his study He randomly assigns 1 class=

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Cricetus

Answer:

The difference throughout the means isn't significant because a difference of 2.5 isn't very likely. A further explanation is provided below.

Step-by-step explanation:

Given:

The mean time of warm up group,

= 13.2

The mean time of non warm up group,

= 10.7

Now,

The mean difference will be:

= [tex]The \ mean \ time \ of \ warm \ up \ group - The \ mean \ time \ of \ non \ warm \ up \ group[/tex]

By substituting the values, we get

= [tex]13.2-10.7[/tex]

= [tex]2.5[/tex]

Thus the above is the appropriate response.

Answer:

C.  The difference in the means is significant because a difference of 2.5 is not very likely.

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