Answer :
The correct option which states the conclusion made from the carter study is [tex]\fbox{\begin\\\ option D\\\end{minispace}}[/tex].
Further explanation:
In the question it is given that thirty track and field athletes volunteered to participate in the study to determine the effect of warming up on the performance of athletes.
To obtain the data for the study carter randomly assign [tex]15[/tex] athletes to the warm up of [tex]10[/tex] minutes and the remaining [tex]15[/tex] athletes didn’t do any warm up.
After the warm up of [tex]15[/tex] athletes all the [tex]30[/tex] athletes were made to sprint for the same distance.
After the sprint it was observed that the mean time for the warm-up group was [tex]10.7[/tex] seconds and the mean time for other group was [tex]13.2[/tex] seconds.
The mean difference of the data obtained is calculated as follows:
[tex]\fbox{\begin\\\ Mean difference=Mean time of warm up group-mean time of non-warm up group\\\end{minispace}}[/tex]
[tex]\fbox{\begin\\\ Mean difference = 13.2-10.7\\\end{minispace}}[/tex]
[tex]\fbox{\begin\\\ Mean difference = 2.5 \\\end{minispace}}[/tex]
The mean differences are plotted in the dot plot in the figure attached below.
The frequency of difference is [tex]2.5[/tex] which is not likely to happen in the graph as it shown in the dot plot.
Option A
The difference in the means is significant because a difference of [tex]2.5[/tex] is very likely.
From the figure attached below it is observed that the difference in the means is not significant because the difference of [tex]2.5[/tex] is not very likely.
So, as per the above statement it is concluded that option A is incorrect.
Option B
The difference in the means is not significant because a difference of [tex]2.5[/tex] is very likely
From the figure attached below it is observed that the difference in the means is not significant because the difference of [tex]2.5[/tex] is not very likely.
So, as per the above statement it is concluded that option B is incorrect.
Option C
The difference in the means is significant because a difference of [tex]2.5[/tex] is not very likely.
From the figure attached below it is observed that the difference in the means is not significant because the difference of [tex]2.5[/tex] is not very likely.
So, as per the above statement it is concluded that option C is incorrect.
Option D
The difference in the means is not significant because a difference of [tex]2.5[/tex] is not very likely.
From the figure attached below it is observed that the difference in the means is not significant because the difference of [tex]2.5[/tex] is not very likely.
So, as per the above statement it is concluded that option D is correct.
Therefore, the correct option which states the conclusion made from the carter study is [tex]\fbox{\begin\\\ option D\\\end{minispace}}[/tex].
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Answer details
Grade: High school
Subject: Mathematics
Chapter: Dot plot
Keywords: Plot, dot plot, statistics, probability, mean, significant, most likely, mean difference, differences, athlete, warm up, very likely,
median, mode, frequency, frequency distribution,
Answer:
The answer is actually C. here's proof:
It also makes sense since 2.5 is very unlikely, so the change is significant.
