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valentina_108 [34]
4 years ago
7

An athletic coach conducted an experiment to test whether a four week strength training program will reduce the number of muscul

ar injuries that occur during athletic events. The coach randomly selected 30 athletes from several sports and assigned 15 athletes to a four week strength training program. The remaining 15 athletes did not participate in any type of strength training program during the four weeks of the program. After the program was completed, the coach monitored each of the 30 athletes for five athletic events. At the end of this process, he reported that the average number of muscular injuries for athletes enrolled in the strength training program is equal to the average number of muscular injuries for athletes not enrolled in the strength training program. What can be concluded from the coach's report? A. There is not enough information to make any conclusions regarding the coach's report. B. It can be concluded that the strength training program does not reduce the number of muscular injuries that occur during an athletic event. C. It can be concluded that the strength training program increases the number of muscular injuries that occur during an athletic event. D. It can be concluded that the strength training program reduces the number of muscular injuries that occur during an athletic event.
Mathematics
1 answer:
Butoxors [25]4 years ago
8 0

Answer: B. It can be concluded that the strength training program does not reduce the number of muscular injuries that occur during an athletic event.

Step-by-step explanation: In the question it states the average number of muscular injuries for athletes enrolled in the strength training program is equal to the average number of muscular injuries for athletes not enrolled in the strength training program.

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Answer:

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

Step-by-step explanation:

Data given and notation  

\bar X=1.6 represent the sample mean

s=0.46 represent the sample deviation

n=32 sample size  

\mu_o =1.35 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:  

Null hypothesis:\mu =1.35  

Alternative hypothesis:\mu \neq 1.35  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

P-value  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

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Answer: 5.25

Step-by-step explanation: .35*15=5.25

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