Question

You wish to test the following claim (HaHa) at a significance level of α=0.10α=0.10.

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

You obtain the following two samples of data.

Sample #1

Sample #2

78.9	58	49.9	49.9
88.8	26.3	40.5	73.2
36.9	56	92.7	57.5
50.5	55.5	78.9	51.7
47.9	61.3	68.3	57.5
71.2	40.5	77.7	71.7
56	43.4	98.7	44.2
49.9	43.4	63.6	57.5
92.7	35.5	67.3	66.9
101.9	40.5	92.7	81.7
60.9	73.7	84.8	33.8
74.8	45.8

23.8	84	57.2	79
38	58.6	57.2	43.1
79	14.7	79	72.6
84	66.8	45.1	77.7
26.5	66.8	59.6	71
57.7	25.2	53	26.5
41.5	57.2	54.4	63.8
55.8	61.1	36.8	64.4
53	76.5	36.8	73.5
53.9	49.8	53	43.6
14.7	39.8	56.2	23.8
34	49.4	41.5	54.4
72.6	68.8

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 =

Answer 1

For Sample 1 :

∑x = 2851

∑x² = 192739

n1 = 46

Mean , x̅1 = Ʃx/n = 2851/46 = 61.9783

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(192738.52-(2851)²/46)/(46-1)] = 18.8789

For Sample 2 :

∑x = 2676.4

∑x² = 159627

n2 = 50

Mean , x̅2 = Ʃx/n = 2676.4/50 = 53.5280

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(159626.78-(2676.4)²/50)/(50-1)] = 18.2748

--

Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 > µ2

Test statistic:

t = (x̅1 - x̅2)/√(s1²/n1 + s2²/n2) = (61.9783 - 53.528)/√(18.8789²/46 + 18.2748²/50) = 2.225

df = ((s1²/n1 + s2²/n2)²)/[(s1²/n1)²/(n1-1) + (s2²/n2)²/(n2-1) ] = 92.7363 = 93

p-value = T.DIST.RT(2.2247, 93) = 0.0143

Decision:

p-value < α, Reject the null hypothesis