Question

In: Statistics and Probability

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 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.


Solutions

Expert Solution

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.


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