HELP! PLEASE!! Simplify the expression.
1 answer:
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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>
Answer: 0.7257
Step-by-step explanation:
Given : The weights of steers in a herd are distributed normally.
Standard deviation :
Let x be the weight of the randomly selected steer .
Z-score :
The the probability that the weight of a randomly selected steer is greater than 920 lbs using standardized normal distribution table :
Hence, the probability that the weight of a randomly selected steer is greater than 920lbs =0.7257