H 0 : μ ⩾ 60 H 1 : μ < 60 x ˉ = 62 + 62 + 68 + 48 + 51 + 60 + 51 + 57 + 57 + 41 + 62 + 50 + 53 + 34 + 62 + 61 16 = 54.9375 s 2 = 6 2 2 + 6 2 2 + 6 8 2 + 4 8 2 + 5 1 2 + 6 0 2 + 5 1 2 + 5 7 2 + 5 7 2 + 4 1 2 + 6 2 2 + 5 0 2 + 5 3 2 + 3 4 2 + 6 2 2 + 6 1 2 − 16 ⋅ 54.937 5 2 16 − 1 = 78.7292 s = s 2 = 78.7292 = 8.87295 T = n x ˉ − μ s = 16 54.9375 − 60 8.87295 = − 2.28222 t n − 1 = t 15 P − v a l u e : P ( T < − 2.28222 ) = F T , 15 ( − 2.28222 ) = 0.0187 H_0:\mu \geqslant 60\\H_1:\mu <60\\\bar{x}=\frac{62+62+68+48+51+60+51+57+57+41+62+50+53+34+62+61}{16}=54.9375\\s^2=\frac{62^2+62^2+68^2+48^2+51^2+60^2+51^2+57^2+57^2+41^2+62^2+50^2+53^2+34^2+62^2+61^2-16\cdot 54.9375^2}{16-1}=78.7292\\s=\sqrt{s^2}=\sqrt{78.7292}=8.87295\\T=\sqrt{n}\frac{\bar{x}-\mu}{s}=\sqrt{16}\frac{54.9375-60}{8.87295}=-2.28222~t_{n-1}=t_{15}\\P-value:\\P\left( T<-2.28222 \right) =F_{T,15}\left( -2.28222 \right) =0.0187 H 0 : μ ⩾ 60 H 1 : μ < 60 x ˉ = 16 62 + 62 + 68 + 48 + 51 + 60 + 51 + 57 + 57 + 41 + 62 + 50 + 53 + 34 + 62 + 61 = 54.9375 s 2 = 16 − 1 6 2 2 + 6 2 2 + 6 8 2 + 4 8 2 + 5 1 2 + 6 0 2 + 5 1 2 + 5 7 2 + 5 7 2 + 4 1 2 + 6 2 2 + 5 0 2 + 5 3 2 + 3 4 2 + 6 2 2 + 6 1 2 − 16 ⋅ 54.937 5 2 = 78.7292 s = s 2 = 78.7292 = 8.87295 T = n s x ˉ − μ = 16 8.87295 54.9375 − 60 = − 2.28222 t n − 1 = t 15 P − v a l u e : P ( T < − 2.28222 ) = F T , 15 ( − 2.28222 ) = 0.0187
Since the P-value is less than the significance level, the null hypothesis is declined. The mean value is less than 60.
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