Please show manually step by step all work in deriving your answer for the next

Question

Please show manually step by step all work in deriving your answer for the next 5 operations (no software)

North University (NU) wants to determine if its students spend more time studying than do students attending Busse University (BU).   Random samples of 9 students are taken at BU and NU. The two samples reveal the following data:

Daily hours studying at NU 6 3 0 3 6 0 2 3 4

Daily hours studying at BU 2 2 3 2 3 2 1 1 2

At the .05 level of significance, is there evidence a NU student spends more time studying on average than a BU student? (Assume both populations of student time spent studying follow a normal distribution and have equal variances.)

Explanation / Answer

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   <=   0  
Ha:   u1 - u2   >   0  

At level of significance =    0.05          

As we can see, this is a    right   tailed test.      
Calculating the means of each group,              
              
X1 =    3          
X2 =    2          
              
Calculating the standard deviations of each group,              
              
s1 =    2.179449472          
s2 =    0.707106781          
              
Thus, the pooled standard deviation is given by              
              
S = sqrt[((n1 - 1)s1^2 + (n2 - 1)(s2^2))/(n1 + n2 - 2)]               
              
As n1 =    9   , n2 =    9  
              
Then              
              
S =    1.620185175          
              
Thus, the standard error of the difference is              
              
Sd = S sqrt (1/n1 + 1/n2) =    0.763762616          
              
As ud = the hypothesized difference between means =    0   , then      
              
t = [X1 - X2 - ud]/Sd =    1.309307341          
              
Getting the critical value using table/technology,              
df = n1 + n2 - 2 =    16          
tcrit =    +   1.745883676      
              
Getting the p value using technology,              
              
p =    0.104460949          
              
As t < 1.746, and P > 0.05, we   FAIL TO REJECT THE NULL HYPOTHESIS.          

Hence, there is no significant evidence that an NU student spend more time studying on average than a BU student. [CONCLUSION]

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Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!

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