(-∞-)help me help me help me help me help me help me help me help me help me help me help me help me help me help me help me hel
p me help me help me
2 answers:
You might be interested in
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:
<u><em>Step( i )</em></u>:-
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.
<u><em> Null hypothesis : H₀ : μ = 5.5</em></u>
<u><em> Alternative Hypothesis : H₁ : μ < 5.5</em></u>
Level of significance ∝ = 0.01
<u><em>Step(ii)</em></u>:-
The test statistic
On calculation , we get
t = -3.33
|t| = |-3.33| = 3.33
<u><em>Step(iii)</em></u>:-
<u><em>P - value</em></u>
<u><em>The degrees of freedom γ = n-1 = 16-1 =15</em></u>
The calculated value t = 3.33 (check t-table) lies between the 0.001 to 0.005
0.001 < P < 0.005
<u>Condition(i)</u>
P - value < ∝ then reject H₀
<u>Condition(ii)</u>
P - value > ∝ then Accept H₀
we observe that 0.001 < P < 0.005
P- value < 0.01
we rejected H₀
<em>(or)</em>
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
<u><em>Conclusion</em></u>:-
Accepted Alternative hypothesis H₁
The Claim that the true average is smaller than 5.5
<u><em></em></u>
You write the equation like this:
But if you want to solve it then you do it like this:
But if you want to solve it then you do it like this: