Answer
given,
mean = 12 Kg
standard deviation = 0.5 Kg
assume the observed statistic is = 11.1
now, [tex]\bar{X}=11.1 , \mu = 12 , \sigma = 0.5[/tex]
assuming the number of sample = 4
n = 4
Hypothesis test:
H₀ : μ≥ 12
Ha : μ < 12
now,
significant level α = 0.05
[tex]z* = \dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
[tex]z* = \dfrac{11.1-12}{\dfrac{0.5}{\sqrt{4}}}[/tex]
z* = -3.60
Test statistics, Z* = -3.60
P-value
P(Z<-3.60) = 0.002 (from z- table)
P- value = 0.002
now,
reject the value of H₀ when P-value < α
0.002 < 0.05
since, it is less P-value < α , we have to reject the null hypothesis
