The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar businesses reported the mean number of customers waiting to have their oil changed on Saturday morning is 3.6. Suppose the local oil changing business owner wants to perform a hypothesis test. The null hypothesis is the population mean is 3.6 and the alternative hypothesis that the population mean is not equal to 3.6. The owner took a random sample of 16 Saturday mornings during the past year and determined the sample mean is 4.2 and the sample standard deviation is 1.4. It can be assumed that the population is normally distributed and the level of significance is 0.05. The critical t value for this problem is _______.
Answers
Answer:
2.131
Step-by-step explanation:
Given : Sample size = 16
Sample mean is 4.2
Sample standard deviation is 1.4.
Level of significance is 0.05.
To Find : critical t value
Solution:
Sample size = 16
Since n < 30
So we will use t - test
We are supposed to find critical t value
Degree of freedom = n-1 = 16-1 = 15
Level of significance =α= 0.05
Now refer the t table for t critical
=
= 2.131
Hence The critical t value for this problem is 2.131
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Answers
Step-by-step explanation:
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Answers
Answer:
C(60) = 2.7*10⁻⁴
t = 1870.72 s
Step-by-step explanation:
Let x(t) be the amount of chlorine in the pool at time t. Then the concentration of chlorine is
C(t) = 3*10⁻⁴*x(t).
The input rate is 6*(0.001/100) = 6*10⁻⁵.
The output rate is 6*C(t) = 6*(3*10⁻⁴*x(t)) = 18*10⁻⁴*x(t)
The initial condition is x(0) = C(0)*10⁴/3 = (0.03/100)*10⁴/3 = 1.
The problem is to find C(60) in percents and to find t such that 3*10⁻⁴*x(t) = 0.002/100.
Remember, 1 h = 60 minutes. The initial value problem is
dx/dt= 6*10⁻⁵ - 18*10⁻⁴x = - 6* 10⁻⁴*(3x - 10⁻¹) x(0) = 1.
The equation is separable. It can be rewritten as dx/(3x - 10⁻¹) = -6*10⁻⁴dt.
The integration of both sides gives us
Ln |3x - 0.1| / 3 = -6*10⁻⁴*t + C or |3x - 0.1| = e∧(3C)*e∧(-18*10⁻⁴t).
Therefore, 3x - 0.1 = C₁*e∧(-18*10⁻⁴t).
Plug in the initial condition t = 0, x = 1 to obtain C₁ = 2.9.
Thus the solution to the IVP is
x(t) = (1/3)(2.9*e∧(-18*10⁻⁴t)+0.1)
then
C(t) = 3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴t)+0.1)
If t = 60
We have
C(60) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴*60)+0.1) = 2.7*10⁻⁴
Now, we obtain t such that 3*10⁻⁴*x(t) = 2*10⁻⁵
3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 2*10⁻⁵
t = 1870.72 s