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 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 =

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

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