An engineer in charge of water rationing for the U.S. Army wants to determine if

Question

An engineer in charge of water rationing for the U.S. Army wants to determine if the average male soldier spends less time in the shower than the average female soldier. Let mrepresent the average time in the shower of male soldiers and f represent the average time in the shower of female soldiers.

b) Among a sample of 54 male soldiers the average shower time was found to be 2.9 minutes and the standard deviation was found to be 0.65 minutes. Among a sample of 61 female soldiers the average shower time was found to be 3.31 minutes and the standard deviation was found to be 0.71 minutes. What is the test statistic? Give your answer to three decimal places.  

c) What is the P-value for the test? Give your answer to four decimal places.  

d) Using a 0.01 level of significance, what is the appropriate conclusion?

Conclude that the average shower time for males is equal to the average shower time for females because the P-value is less than 0.01.

Reject the claim that the average shower times are different for male and female soldiers because the P-value is greater than 0.01.    

Fail to reject the claim that the average shower times are the same for male and female soldiers because the P-value is greater than 0.01.

Conclude that the average shower time for males is less than the average shower time for females because the P-value is less than 0.01.

Explanation / Answer

b)

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   >=   0  
Ha:   u1 - u2   <   0  
At level of significance =    0.01          
As we can see, this is a    left   tailed test.      
Calculating the means of each group,              
              
X1 =    2.9          
X2 =    3.31          
              
Calculating the standard deviations of each group,              
              
s1 =    0.65          
s2 =    0.71          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    54          
n2 = sample size of group 2 =    61          

Also, sD =    0.126838513          
              
Thus, the z statistic will be              
              
z = [X1 - X2 - uD]/sD =    -3.232456679   [ANSWER, TEST STATISTIC]

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c)

As this is left tailed, then the P value is

P = 0.000613654 [ANSWER]

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d)

As P < 0.01, we reject Ho. Hence,

OPTION D: Conclude that the average shower time for males is less than the average shower time for females because the P-value is less than 0.01. [ANSWER]

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Hi! I used z distirbution because both sample sizes are greater than 30. If you use another method to solve this, please resubmit this question together indicating which method you want to use. That way we can continue helping you! Thanks!

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