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faust18 [17]
4 years ago
10

A running coach wants to know if participating in weekly running clubs significantly improves the time to run a mile. The runnin

g coach collects mean mile times for 42 runners who participate in weekly running clubs with 54 runners who do not run in clubs. The running coach measures times in January and June of the same year. All times are in seconds, and the runners all started with mile times between 8 minutes (480 seconds) and 9 minutes (540 seconds). Here are the results:
January June Change
Mean (seconds) StDev (seconds) Mean (seconds) StDev (seconds) Mean (seconds) StDev (seconds)
Running Club N = 42 530 28 520 23 10 4
Not in a Club N = 54 527 25 519 22 8 3
What is the best method for the running coach to use to determine if participating in weekly running clubs significantly improves the time to run a mile?
A) Use the difference in sample means (530 – 520) in a hypothesis test for a difference in two population means.
B) Use the difference in sample means (10 and 8) in a hypothesis test for a difference in two population means.
C) Use the sample mean 10 in a hypothesis test for a population mean.

Mathematics
1 answer:
patriot [66]4 years ago
7 0

Answer:

Option B is correct.

Use the difference in sample means (10 and 8) in a hypothesis test for a difference in two population means.

Step-by-step Explanation:

The clear, complete table For this question is presented in the attached image to this solution.

It should be noted that For this question, the running coach wants to test if participating in weekly running clubs significantly improves the time to run a mile.

In the data setup, the mean time to run a mile in January for those that participate in weekly running clubs and those that do not was provided.

The mean time to run a mile in June too is provided for those that participate in weekly running clubs and those that do not.

Then the difference in the mean time to run a mile in January and June for the two classes (those that participate in weekly running clubs and those that do not) is also provided.

Since, the aim of the running coach is to test if participating in weekly running clubs significantly improves the time to run a mile, so, it is logical that it is the improvements in running times for the two groups that should be compared.

Hence, we should use the difference in sample means (10 and 8) in a hypothesis test for a difference in two population means.

Hope this Helps!!!

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The coaches at Xavier Elementary School bought cases of sports drinks for a field day. They bought 76 cases of drinks. Each case
solong [7]

Answer:

608

Step-by-step explanation:

There are a total of 1,824 drinks since there were 76 cases at 24 drinks a piece. This creates a total number of 24*76 = 1824.

If each student receives 3 then 1824/3 = 608. There were 608 students.

7 0
3 years ago
A little monkey had some peaches. On the first day he decided to keep ½ of his peaches. He gave the rest away. Then he ate one.
Hunter-Best [27]

Answer:

22

Step-by-step explanation:

Working backwards

1+1 then x2 is 4

4+1 then x2 is 10

10+1 then x2 is 22

8 0
2 years ago
As part of a drug trial, a survey was done to gather information about the participants' experiences with side effects. Here are
Paraphin [41]

The Venn diagram which represents the distribution of the participant in the drug trial is attached below. The Number of participants in the drug trial that has anxiety is 370

We can find the number of participants who has dizziness(D). Fatigue(F) and anxiety(A) can be calculated thus :

n(D) = n(D only) + (DnF only) + (DnA only) + (DnAnF)

271 = 36 + 86 + 23 + x

271 = 145 + x

x = 271 - 145

x = 126

  • n(DnAnF) = 126

<u>Number who has </u><u>atleast one of the three</u><u> side effects can be expressed thus</u> :

n(A only) + n(F only) + n(D only) + (DnF only) + (DnA only) + (DnAnF) + n(FnA only) = 585

36 + 23 + 62 + 86 + 126 + 93 + n(FnA) only = 585

  • n(FnA only) = 585 - 426 = 159

<u>Number of participants</u><u> who has </u><u>anxiety</u><u> can be calculated thus</u> :

n(DnAnF) + (DnA only) + n(FnA only) + n(A only)

126 + 23 + 159 + 62 = 370 participants

Therefore, 370 of the total participants has Anxiety.

Learn more : brainly.com/question/12570490

8 0
2 years ago
Please help me with this and I will give one of the answers most brainliest
inna [77]
A)

A=8\qquad\qquad B=12.5\\\\\\d(A,B)=|12.5-8|=|4.5|=\boxed{4.5\,\text{cm}}

b)

C=?\\\\\\d(C,A)=2\cdot d(C,B)\\\\|A-C|=2\cdot|B-C|\\\\|8-C|=2\cdot|12.5-C|\\\\&#10;|8-C|=|25-2C|\\\\\\8-C=25-2C\qquad\vee\qquad8-C=-(25-2C)\\\\-C+2C=25-8\qquad\vee\qquad8-C=-25+2C\\\\C=17\qquad\vee\qquad8+25=2C+C\\\\C=17\qquad\vee\qquad33=3C\quad|:3\\\\\boxed{C=17\qquad\vee\qquad C=11}

Mark 11 or 17.
4 0
3 years ago
What is 7 over 9 subtract 1 over 3
Firlakuza [10]
The answer is 4/9 or 0.44444
5 0
4 years ago
Read 2 more answers
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