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Answer:
The 95% t-confidence interval for the difference in mean is approximately (-2.61, 1.16), therefore, there is not enough statistical evidence to show that there is a change in waiting time, therefore;
The change in the call waiting time is not statistically significant
Step-by-step explanation:
The given call waiting times are;
24.16, 20.17, 14.60, 19.79, 20.02, 14.60, 21.84, 21.45, 16.23, 19.60, 17.64, 16.53, 17.93, 22.81, 18.05, 16.36, 15.16, 19.24, 18.84, 20.77
19.81, 18.39, 24.34, 22.63, 20.20, 23.35, 16.21, 21.73, 17.18, 18.98, 19.35, 18.41, 20.57, 13.00, 17.25, 21.32, 23.29, 22.09, 12.88, 19.27
From the data we have;
The mean waiting time before the downsize, = 18.7895
The mean waiting time before the downsize, s₁ = 2.705152
The sample size for the before the downsize, n₁ = 20
The mean waiting time after the downsize, = 19.5125
The mean waiting time after the downsize, s₂ = 3.155945
The sample size for the after the downsize, n₂ = 20
The degrees of freedom, df = n₁ + n₂ - 2 = 20 + 20 - 2 = 38
df = 38
At 95% significance level, using a graphing calculator, we have; = ±2.026192
The t-confidence interval is given as follows;
Therefore;
(18.7895 - 19.5125) - 2.026192*(2.705152²/20 + 3.155945²/20)^(0.5)
The 95% CI = -2.6063 < μ₂ - μ₁ < 1.16025996668
By approximation, we have;
The 95% CI = -2.61 < μ₂ - μ₁ < 1.16
Given that the 95% confidence interval ranges from a positive to a negative value, we are 95% sure that the confidence interval includes '0', therefore, there is sufficient evidence that there is no difference between the two means, and the change in call waiting time is not statistically significant.
Answer:
x = 2 , y = 0 , z = 3
Step-by-step explanation:
Cramer's rule is a rule through which we can find the solution of linear equation.
we have the three linear equations as
x+2y+3z=11
2x+y+2z=10
3x+2y+z=9
AX=B
A: coefficient matrix
X= unknown vectors(x,y,z)
D = values of the linear equation (11 , 10 , 9)
now we find the determinant of the given linear equation
determinant of the matrix will be
A = = 1(1-4) - 2(2-6) + 3(4 - 3)
= 1(-3) - 2(-4) + 3(1)
= -3+8+3 = 8
also D
so the determinant is Non zero we can apply Cramer's rule
we will be replacing the first column of the coefficient matrix A with the values of D
by replacing the first column we will get the value of the variable 'x'
Dx = = 11(1-4) -2(10-18) + 3(20-9) = -33+16+33 = 16
x = =
= 2
similarly
Dy = = 1(10-18) -11(2-6) + 3(18 -30) = -8 +44 -36 = 0
y = = 0
Dz= = 1(9 - 20) -2(18 - 30) + 11(4 -3) = -11 +24 +11 = 24
z = =
so we have the solution as
x = 2 , y = 0 , z = 3
Therefore the solution for the given linear equations is (2,0,3).