In this case,
S a m p l e m e a n = x ‾ = 16.45 Sample\:mean=\overline{x}=16.45 S am pl e m e an = x = 16.45
S a m p l e s t a n d a r d d e v i a t i o n = s = 3.59 Sample\:standard\:deviation=s=3.59\: S am pl e s t an d a r d d e v ia t i o n = s = 3.59
S a m p l e s i z e = n = 20 Sample\:size=n=20 S am pl e s i ze = n = 20
D e g r e e s o f f r e e d o m = 19 \:Degrees\:of\:freedom=19 De g rees o f f ree d o m = 19
S i g n i f i c a n c e l e v e l = α = 0.05 Significance\:level=\alpha =0.05 S i g ni f i c an ce l e v e l = α = 0.05
Step 1: H 0 : μ = 16 H_0:\mu=16 H 0 : μ = 16
H 1 : μ ≠ 16 H_1:\mu\:\:\not= \:16 H 1 : μ = 16 ( T w o (Two ( Tw o t a i l e d tailed t ai l e d t e s t ) test) t es t )
Step 2: T h e t e s t s t a t i s t i c i s , t = x ‾ − μ ( s n ) The\:test\:statistic\:is,\:t=\frac{\overline{x}-\mu }{\left(\frac{s}{\sqrt{n}}\right)}\: T h e t es t s t a t i s t i c i s , t = ( n s ) x − μ
t = 16.45 − 16 ( 3.59 20 ) = 0.5606 t=\frac{16.45-16}{\left(\frac{3.59}{\sqrt{20}}\right)}=0.5606 t = ( 20 3.59 ) 16.45 − 16 = 0.5606
Step 3: P − v a l u e = P ( ∣ t ∣ > 0.5606 ) = 0.5816 P-value=P\left(\left|t\right|>0.5606\right)=0.5816 P − v a l u e = P ( ∣ t ∣ > 0.5606 ) = 0.5816
Step 4: Conclusion: Since the P − v a l u e P-value P − v a l u e is greater than the significance level, we fail to reject H o H_o H o .
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