Answer:
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Answer:
The degrees of freedom associated with the critical value is 25.
Step-by-step explanation:
The number of values in the final calculation of a statistic that are free to vary is referred to as the degrees of freedom. That is, it is the number of independent ways by which a dynamic system can move, without disrupting any constraint imposed on it.
The degrees of freedom for the t-distribution is obtained by substituting the values of n1 and n2 in the degrees of freedom formula.
Degrees of freedom, df = n1+n2−2
= 15+12−2=27−2=25
Therefore, the degrees of freedom associated with the critical value is 25.
Given:
The given system of equations is:
To find:
The solution to this system of equations by graphing.
Solution:
We have,
The table of values for first equation is:
x y
0 1
1 -1
Plot the points (0,1) and (1,-1) on a coordinate plane and connect them a straight line.
The table of values for second equation is:
x y
0 -4
2 -3
Plot the points (0,-4) and (2,-3) on a coordinate plane and connect them a straight line.
The graphs of given equations are shown in the below figure.
From the below figure, it is clear that the lines intersect each other at point (2,-3). So, the solution of the given system of equations is (2,-3).
Therefore, the solution to this system of equations is:
x-coordinate: 2
y-coordinate: -3
It is already in standard form.
Answer:
We accept H₀. We do not have argument to keep the claim that the mean breaking strength has increased
Step-by-step explanation:
Normal Distribution
Population Mean μ₀ = 1850 pounds
Standard Deviation σ = 90 pounds
Type of test
Null Hypothesis H₀ ⇒ μ = μ₀
Alternative Hypothesis Hₐ ⇒ μ > μ₀
A one tail test (right)
n = 21 as n < 30 we use t-student table
degree of fredom 20
t = 2.845
Sample mean μ = 1893
Then, we compute t statistics
t(s) = [ 1893 - 1850 ] / 90/ √n
t(s) = 43 * 4,583 / 90
t(s) = 197,069 / 90
t(s) = 2,190
And we compare t and t(s)
t(s) = 2.190
t = 2.845
Then t(s) < t
We are in the acceptance zone, we accept H₀
