Question

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

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

You obtain the following two samples of data.

sample 1

66.9	49.9	58.3	61.6
29.4	78	40.8	43.1
56.4	61.1	85.3	68.4
70	58.8	53	78
51.5	55.9	43.1	45.8
16.7	76.6	60.7	91
63.5	65.4	64.5	62.5
92.9	46.4	58.8	71.7
70	103.7	55.9	69.5
61.6	83.1	60.7	79.6
85.3	72.2	59.3	63
81.3

Sample 2

52.9	59.8	47.2	80.7
68.4	49.8	25.6	49.3
87.8	29.8	99.6	47.8
32.3	46.7	58.7	57.1
36.1	67	62.1	20
57.6	44.6	56	20
35.4	34.6	82.2	28.8
45.1	32.3	55	29.8
37.4	25.6	24.4	59.2
56.6	99.6	49.3	46.2
45.1	68.4	36.1	53.4
58.7	21.6	60.4	42.3
52.4	59.2	35.4	45.1
60.4

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 complicated formula for degrees of freedom in Worksheet 7.3. (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 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

For Sample 1 :       
∑x =    2871.2  
∑x² =    195264.48  
n1 =    45  
Mean , x̅1 = Ʃx/n =    2871.2/45 =    63.8044
Standard deviation, s1 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] =    √[(195264.48-(2871.2)²/45)/(45-1)] =    16.5620
For Sample 2 :       
∑x =    2636.9  
∑x² =    149255.41  
n2 =    53  
Mean , x̅2 = Ʃx/n =    2636.9/53 =    49.7528
Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] =    √[(149255.41-(2636.9)²/53)/(53-1)] =    18.6373
--

Null and Alternative hypothesis:  
Ho : µ1 = µ2  
H1 : µ1 ≠ µ2  
Test statistic:  
t = (x̅1 - x̅2)/√(s1²/n1 + s2²/n2) = (63.8044 - 49.7528)/√(16.562²/45 + 18.6373²/53) =    3.951
df = ((s1²/n1 + s2²/n2)²)/[(s1²/n1)²/(n1-1) + (s2²/n2)²/(n2-1) ] = 95.786 = 96  
p-value =T.DIST.2T(3.9509, 96) =    0.0001
Decision:  
p-value < α, Reject the null hypothesis  
Conclusion:  
There is sufficient evidence to warrant rejection of the claim that the first population mean is not equal to the second population mean.