Question

You wish to test the following claim (HaHa) at a significance level of α=0.005

Ho:μ1=μ2
Ha:μ1≠μ2

You believe both populations are normally distributed, but you do not know the standard deviations for either. And you have no reason to believe the variances of the two populations are equal You obtain a sample of size n1=12 with a mean of ¯x1=76.2x and a standard deviation of s1=14.6 from the first population. You obtain a sample of size n2=25 with a mean of ¯x2=92.3x and a standard deviation of s2=5.3 from the second population.

1. What is the test statistic for this sample?
   
   test statistic =  Round to 3 decimal places.
   
2. What is the p-value for this sample? For this calculation, use .
   
   p-value =  Use Technology Round to 4 decimal places.
   
3. The p-value is...
   • less than (or equal to) αα
   • greater than αα
   
   
4. This test statistic leads to a decision to...
   • reject the null
   • accept the null
   • fail to reject the null
   
   
5. As such, the final conclusion is that...
   • There is sufficient evidence to warrant rejection of the claim that the first population mean is not equal to the second population mean.
   • There is not sufficient evidence to warrant rejection of the claim that the first population mean is not equal to the second population mean.
   • The sample data support the claim that the first population mean is not equal to the second population mean.
   • There is not sufficient sample evidence to support the claim that the first population mean is not equal to the second population mean.

Answer 1

Ho :   µ1 - µ2 =   0          
Ha :   µ1-µ2 ╪   0          
                  
Level of Significance ,    α =    0.05          
                  
Sample #1   ---->   sample 1          
mean of sample 1,    x̅1=   76.20          
standard deviation of sample 1,   s1 =    14.6          
size of sample 1,    n1=   12          
                  
Sample #2   ---->   sample 2          
mean of sample 2,    x̅2=   92.300          
standard deviation of sample 2,   s2 =    5.30          
size of sample 2,    n2=   25          
                  
difference in sample means = x̅1-x̅2 =    76.200   -   92.3000   =   -16.1000
                  
std error , SE =    √(s1²/n1+s2²/n2) =    4.3459          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -16.1000   /   4.3459   ) =   -3.705
                  

                  

p-value =        0.0030   (excel function: =T.DIST.2T(t stat,df) )      
Conclusion:     p-value<α , Reject null hypothesis              

The sample data support the claim that the first population mean is not equal to the second population mean.

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