A random sample of 9 observations from one population revealed a sample mean of

Question

A random sample of 9 observations from one population revealed a sample mean of 25 and a sample standard deviation of 4.0. A random sample of 4 observations from another population revealed a sample mean of 29 and a sample standard deviation of 4.6.

State the decision rule. (Negative amounts should be indicated by a minus sign. Round your answer to 3 decimal places.)

Compute the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)

.

A random sample of 9 observations from one population revealed a sample mean of 25 and a sample standard deviation of 4.0. A random sample of 4 observations from another population revealed a sample mean of 29 and a sample standard deviation of 4.6.

Explanation / Answer

a)

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    two   tailed test.  
Getting the critical value using table/technology,              
df = n1 + n2 - 2 =    11          

Thus,

tcrit =    +/-   3.105806516      

Hence,

The decision rule is to reject H0 if t < -3.106 or t > 3.106. [ANSWER]

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b)

  
Calculating the means of each group,              
              
X1 =    25          
X2 =    29          
              
Calculating the standard deviations of each group,              
              
s1 =    4          
s2 =    4.6          
              
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 =    4  
              
Then              
              
S =    4.172202383          

Hence,

S^2 = 17.40727272 [ANSWER, POOLED POPULATION VARIANCE]

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c)

              
Thus, the standard error of the difference is              
              
Sd = S sqrt (1/n1 + 1/n2) =    2.507181604          
              
As ud = the hypothesized difference between means =    0   , then      
              
t = [X1 - X2 - ud]/Sd =    -1.595416939   [ANSWER, TEST STATISTIC]

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d)      
              

As |t| < 3.106, we FAIL TO REJECT HO. [ANSWER]

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e)
              
Getting the p value using technology, as this is two tailed,              
              
p =    0.138925587   [ANSWER, P VALUE]

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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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