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{Exercise 10.17} a. State the null and alternative hypotheses. : mu1 - mu2 is le

ID: 3125380 • Letter: #

Question

{Exercise 10.17}

a. State the null and alternative hypotheses.
: mu1 - mu2 is less than 0- Select your answer -mu1 - mu2 is greater than 0mu1 - mu2 is greater than or equal to 0mu1 - mu2 is equal to 0mu1 - mu2 is not equal to 0mu1 - mu2 is less than 0mu1 - mu2 is less than or equal to 0Item 1  
:mu1 - mu2 is greater than or equal to 0- Select your answer -mu1 - mu2 is greater than 0mu1 - mu2 is greater than or equal to 0mu1 - mu2 is equal to 0mu1 - mu2 is not equal to 0mu1 - mu2 is less than 0mu1 - mu2 is less than or equal to 0Item 2

b. Compute the value of the test statistic. (to 2 decimals)

c. What is the p-value? (to 4 decimals)
  

d. What is your conclusion?
Reject the null hypothesis- Select your answer -Reject the null hypothesisDo not reject the null hypothesisItem 6

Explanation / Answer

a)

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

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b)

At level of significance =    0.05          
As we can see, this is a    right   tailed test.      
Calculating the means of each group,              
              
X1 =    6.82          
X2 =    6.25          
              
Calculating the standard deviations of each group,              
              
s1 =    0.64          
s2 =    0.75          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    16          
n2 = sample size of group 2 =    10          
Thus, df = n1 + n2 - 2 =    24          
Also, sD =    0.28609439          

uD = hypothesized difference =    0      
              
Thus, the t statistic will be              
              
t = [X1 - X2 - uD]/sD =    1.992349448   [ANSWER]

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c)      
              
              
Also, using p values,              
              
p =    0.02891273 = 0.0289 [ANSWER]

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d)
          
              
As P < 0.05,    WE REJECT THE NULL HYPOTHESIS.      

Hence, there is significant evidence at 0.05 level that the consultant with more experience has the higher population mean service rating. [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!