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# stats - a Null hypothesis states that the mean daily email...

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a) Null hypothesis states that the mean daily email usage of managers is equal or significantly higher than the executives. Alternative hypothesis states that the mean daily email usage of managers is significantly lower than the executives. H 0 : µ M ≥µ E H 1: µ M < µ E A one-tail test as this question is concerned with: if a mean is lower than the other. b) degree of freedom, df = N E + N M - 2 = 26 + 27 – 2 = 51 Critical value of t for df = 51 at 0.05 significance level for one-tail test = 1.675 To reject H 0 if t < -1.675 Mean of Exec : X E =(4.3+3.6+4.6+4.7+5.3+6.1+4.9+6.2+5.3+4.5+5.7+5.1+4.7+5.4+5.8+4.7+5.3+5.1+5.3+5.6+4.7+4.1+4.5+5.3+6.7+6.4) / 26 = 5.150 Mean of Manager : X M =(4.4+2.8+4.4+4.1+3.4+4.2+4.1+3.9+4.3+3.1+3.5+3.1+4.7+4.5+2.8+3.9+4.2+3.6+3.7+3.9+3.6+3.2+3.4+3.7+3.6+3.8+4.8) / 27 = 3.804 Standard deviation of Exec :

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Hours Hours - X E (Hours - X E ) 2 ∑(Hours - X E ) 2 ∑(Hours - X E ) 2 / 25
4.3 -0.85 0.7225 3.6 -1.55 2.4025 4.6 -0.55 0.3025 4.7 -0.45 0.2025 5.3 0.15 0.0225 6.1 0.95 0.9025 4.9 -0.25 0.0625 6.2 1.05 1.1025 5.3 0.15 0.0225 4.5 -0.65 0.4225 5.7 0.55 0.3025 5.1 -0.05 0.0025 4.7 -0.45 0.2025 5.4 0.25 0.0625 5.8 0.65 0.4225 4.7 -0.45 0.2025 5.3 0.15 0.0225 5.1 -0.05 0.0025 5.3 0.15 0.0225 5.6 0.45 0.2025 4.7 -0.45 0.2025 4.1 -1.05 1.1025 4.5 -0.65 0.4225 5.3 0.15 0.0225 6.7 1.55 2.4025 6.4 1.25 1.5625 13.325 0.533 Variance = 0.533 Standard deviation = √0.533 = 0.7301 Standard deviation of Manager :

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Hours Hours – X M (Hours – X M ) 2 ∑(Hours – X M ) 2 ∑(Hours – X M ) 2 / 25
4.4 0.596 0.355216 2.8 -1.004 1.008016 4.4 0.596 0.355216 4.1 0.296 0.087616 3.4 -0.404 0.163216

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