Answer:
a) Null hypothesis : H₀ : μ = 5.5
Alternative Hypothesis : H₁ : μ < 5.5
b) The test statistic
|t| = |-3.33| = 3.33
c) P - value lies between in these intervals
0.001 < P < 0.005
Step-by-step explanation:
Step( i ):-
Given data the Population mean 'μ' = 5.5
The small sample size 'n' = 16
The sample mean (x⁻) = 5.25
Given the percentage of SiO2 in a sample is normally distributed with a sigma of 0.3.
Null hypothesis : H₀ : μ = 5.5
Alternative Hypothesis : H₁ : μ < 5.5
Level of significance ∝ = 0.01
Step(ii):-
The test statistic
[tex]t = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }[/tex]
[tex]t = \frac{5.25 -5.5}{\frac{0.3}{\sqrt{16} } }[/tex]
On calculation , we get
t = -3.33
|t| = |-3.33| = 3.33
Step(iii):-
P - value
The degrees of freedom γ = n-1 = 16-1 =15
The calculated value t = 3.33 (check t-table) lies between the 0.001 to 0.005
0.001 < P < 0.005
Condition(i)
P - value < ∝ then reject H₀
Condition(ii)
P - value > ∝ then Accept H₀
we observe that 0.001 < P < 0.005
P- value < 0.01
we rejected H₀
(or)
The tabulated value = 2.60 at 0.01 level of significance with '15' degrees of freedom
The calculated value t = 3.33 > 2.60 at 0.01 level of significance with '15' degrees of freedom
The null hypothesis is rejected
Conclusion:-
Accepted Alternative hypothesis H₁
The Claim that the true average is smaller than 5.5
