Ben is taking a typing class, and the following line graph represents the number of words per minute he was able to type each we
ek. According to the graph, how many more words per minute could Ben type in week four than in week 2? DOK 2
1 answer:
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Answer:
its a really bad graph but i tried
Step-by-step explanation:
3/4x= (3,4)
-1/2= y-1/2 of one
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
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