Answer:
We use students' t distribution therefore degrees of freedom is v= n-2
Step-by-step explanation:
<u>Confidence Interval Estimate of Population Regression Co efficient β.</u>
To construct the confidence interval for β, the population regression co efficient , we use b, the sample estimate of β. The sampling distribution of b is normally distributed with mean β and a standard deviation σ.y.x / √(x-x`)². That is the variable z = b - β/σ.y.x / √(x-x`)² is a standard normal variable. But σ.y.x is not known so we use S.y.x and also student's t distribution rather than normal distribution.
t= b - β/S.y.x / √(x-x`)² = b - β/Sb [Sb = S.y.x / √(x-x`)²]
with v= n-2 degrees of freedom.
Consequently
P [ - t α/2< b - β/Sb < t α/2] = 1- α
or
P [ b- t α/2 Sb< β < b+ t α/2 Sb] = 1- α
Hence a 100( 1-α) percent confidence for β the population regression coefficient for a particular sample size n <30 is given by
b± t α/2 Sb
Using the same statistic a confidence interval for α can be constructed in the same way for β replacing a with b and Sa with Sb.
a± t α/2 Sa
Using the t statistic we may construct the confidence interval for U.y.x for the given value X0 in the same manner
Y~0 ± t α/2(n-2) SY~
Y~0= a+b X0
-3 (-4 + -x ) = (-4(x) × 2 ) + 6
-3 (-4-x) = (-4x × 2) + 6
-12 + 3x = -8x + 6
3x - 12 = 6 - 8x
add 8x on both sides
11x - 12 = 6
add 12 on both sides
11x = 18
<em><u>x</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>1</u></em><em><u>1</u></em><em><u>/</u></em><em><u>1</u></em><em><u>8</u></em>
A square has all of sides equals. So in the bottom of box you have a square 2x2 and the formula for volumen is width*length*thicknes, from the data 2*2*t=200 cubic inches, solve for t=50 inches. Now for an open box you have 4 pieces around the square in the bottom, you add 50+2+50=102 inches, for lenght and width 102inches
Answer:
True, it is Known as a neutral integer Because it is neither negative or positive whole number
Step-by-step explanation:
Explication étape par étape:
Compte tenu des expressions ';
A = 4 (x + 5) -8
B = x² + 15
Nous devons vérifier si A = B pour les deux valeurs de x à x = 1 et x = 3
à quand x = 1
A = 4 (1 + 5) -8
A = 4 (6) - 8
A = 24-8
A = 16
B = x² + 15
B = 1² + 15
B = 1 + 15
B = 16
Donc à quand x = 1, A = B = 16
quand x = 3
A = 4 (x + 5) -8
A = 4 (3 + 5) -8
A = 4 (8) - 8
A = 32-8
A = 24
B = 3² + 15
B = 9 + 15
B = 24
Également lorsque x = 3, A = B = 24
Cela montre que le postulat d'Emmas est juste.
