Answer:
(a) z-value = -2.12
(b) p-value = 0.0170
(c) Reject H0
Step-by-step explanation:
(a)
[tex]std-err=\frac{std-dev}{\sqrt{n}}=\frac{2}{\sqrt{50}}=0.2828[/tex]
[tex]z-value=\frac{X-mean}{std-err}=\frac{19.4-20}{0.2828}=-2.12[/tex]
(b)
with
z-value = -2.12
significance level = 0.05
one-tail hypothesis (H0: μ ≥ 20)
We can see on the normal distribution table (Z-Score table) that
p-value = 0.0170
(c)
Since p value (0.0170) is less than α=0.05 we reject H0
Hope this helps!
