The Answer is X ˉ \bar { X } X ˉ = 3.44,s = .57,and n = 13 at a significance level of .05.Assume population normality. Reject H 0 3 \\t = \frac { 3.44 - 3.0 } { \frac { .57 } { \sqrt { 13 } } } = 2.783 \\t _ { .0512 } = 1.782 \\2.783 > 1.782 , \text { reject } H _ { 0 }\end{array}"> H 0 : μ ≤ 3 H A : μ > 3 t = 3.44 − 3.0 . 57 13 = 2.783 t . 0512 = 1.782 2.783 > 1.782 , reject H 0 \begin{array} { l } H _ { 0 } : \mu \leq 3 \quad H _ { A } : \mu > 3 \\t = \frac { 3.44 - 3.0 } { \frac { .57 } { \sqrt { 13 } } } = 2.783 \\t _ { .0512 } = 1.782 \\2.783 > 1.782 , \text { reject } H _ { 0 }\end{array} H 0 : μ ≤ 3 H A : μ > 3 t = 13 <path d='M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z'/> .57 3.44 − 3.0 = 2.783 t .0512 = 1.782 2.783 > 1.782 , reject H 0

Solved

Test H₀: μ ≤ 3

question 2

Essay

Test H₀: μ ≤ 3.0 versus H_A: μ > 3.0 when

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Definitions:

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A theoretical framework describing the various phases individuals go through during their professional life, each with distinct characteristics, challenges, and opportunities.

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Test H0: μ ≤ 3

Definitions:

Test H₀: μ ≤ 3