Answer
given,
mean = 12 Kg
standard deviation = 0.5 Kg
assume the observed statistic is = 11.1
now,
![\bar{X}=11.1 , \mu = 12 , \sigma = 0.5](https://img.qammunity.org/2021/formulas/mathematics/college/f99ngid0i501st4eq9ihku7unonzhunh1r.png)
assuming the number of sample = 4
n = 4
Hypothesis test:
H₀ : μ≥ 12
Ha : μ < 12
now,
significant level α = 0.05
![z* = \frac{\bar{X}-\mu}{(\sigma)/(√(n))}](https://img.qammunity.org/2021/formulas/mathematics/college/jjx2rg5o6h233s5wavbwsaf8rjrmtv5waz.png)
![z* = (11.1-12)/((0.5)/(√(4)))](https://img.qammunity.org/2021/formulas/mathematics/college/na0zlin2v9mmtftildeefb5bygf21inwtq.png)
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