Answer:
The P-Value is 0.07186
Step-by-step explanation:
GIven that :
Mean = 70
standard deviation = 3.5
sample size n = 49
sample mean = 69.1
The null hypothesis and the alternative hypothesis can be computed as follows;
[tex]H_o : \mu = 70 \\ \\ H_1 : \mu \neq 70[/tex]
The standard z score formula can be expressed as follows;
[tex]\mathtt{z = \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}}}[/tex]
[tex]\mathtt{z = \dfrac{69.1 - 70}{\dfrac{3.5}{\sqrt{49}}}}[/tex]
[tex]\mathtt{z = \dfrac{-0.9}{\dfrac{3.5}{7}}}[/tex]
z = -1.8
Since the test is two tailed and using the Level of significance = 0.05
P- value = 2 × P( Z< - 1.8)
From normal tables
P- value = 2 × (0.03593)
The P-Value is 0.07186
