Question

You wish to test the following claim (HaHa) at a significance level of α=0.05α=0.05.

      Ho:μ1=μ2Ho:μ1=μ2
      Ha:μ1>μ2Ha:μ1>μ2

You obtain the following two samples of data.

Sample #1	Sample #2
67.8 100.5 96.8 59.4 66.7 86.2 89.2 86.7 70.6 100.5 65.8 77.4 78.4 71.9 69.3 61.4 69.6 81.6 90.4 75.3 83.4 70.9 84.5 75.3 62 58.6 77.4 73.8 70.6 83 83 63.6 76.2 61.4 75.6 84.5 92.6 87.6	73.9 77.3 77.8 64.9 67.5 49.7 87 56.5 96.1 75.8 84.1 39.6 63.8 68.5 88.3 84.1 74.4 75.3 95.1 67 57.8 37.1 85.2 103.4 55 79.8 52.5 66.5 28.5 101.9 48.7 75.3 63.8 75.8 28.5 72.9 62.2 97.1 86.4 89.6

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =

What is the p-value for this sample? For this calculation, use the degrees of freedom reported from the technology you are using. (Report answer accurate to four decimal places.)
p-value =

The p-value is...

• less than (or equal to) αα
• greater than αα

This test statistic leads to a decision to...

• reject the null
• accept the null
• fail to reject the null

As such, the final conclusion is that...

• There is sufficient evidence to warrant rejection of the claim that the first population mean is greater than the second population mean.
• There is not sufficient evidence to warrant rejection of the claim that the first population mean is greater than the second population mean.
• The sample data support the claim that the first population mean is greater than the second population mean.
• There is not sufficient sample evidence to support the claim that the first population mean is greater than the second population mean.

Answer 1

Ho :   µ1 - µ2 =   0          
Ha :   µ1-µ2 >   0          
                  
Level of Significance ,    α =    0.05          
                  
Sample #1   ---->   1          
mean of sample 1,    x̅1=   77.09          
standard deviation of sample 1,   s1 =    11.32033286          
size of sample 1,    n1=   38          
                  
Sample #2   ---->   2          
mean of sample 2,    x̅2=   70.868          
standard deviation of sample 2,   s2 =    18.81          
size of sample 2,    n2=   40          
                  
difference in sample means = x̅1-x̅2 =    77.092   -   70.8675   =   6.2246
                  
std error , SE =    √(s1²/n1+s2²/n2) =    3.4958          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   6.2246   /   3.4958   ) =   1.781

------------------------------

p-value =        0.0399   [excel function: =T.DIST.RT(t stat,df) ]

----------------------------------

The p-value is...

• less than (or equal to) αα

------------------

This test statistic leads to a decision to...

• reject the null

-----------------

• The sample data support the claim that the first population mean is greater than the second population mean.