Question

5)Two resistors of resistances R1=100±3 ohm and R2=200±4 ohm are connected

(a) In series, (b) in parallel.

Find the equivalent resistance of the

(a) series combination,

(b) parallel combination.

Use for

(a) the relation R=R1+R2 and for

(b) 1/R=1/R1+1/R2 and ΔR1/R12+ΔR2/R22 ?

Answer 1

(a) Series Combination

⭕️Resistance =R=R1+R2

= R = 100 ± 3 + 200 ± 4 = R = 300 ± 7 ohm

(b) Parallel combination

⭕️Resistance =1/R = 1/R1 + 1/R2 = 1/R = (R1 + R2)/(R1x R2)

 ⇒R = (R1x R2)/(R1+ R2)

 ∴ R = (100 x 200)/(100 + 200)

 ⇒ R = 20000/300 = 200/3

⇒ R = 66.67 ohm

⭕️Now to calculate error:-

 ∴ ∆R/R² = ∆R1/R1²+ ∆R2/R2²

⭕️Putting values we get:-

⇒ ∆R/(66.67)² = 3/(100)²+ 4/(200)²

 ⇒ ∆R/(66.67)²= 3/10000 + 4/40000

 ⇒ ∆R/(66.67)² = 3/10000 + 1/10000

 ⇒ ∆R/(66.67)² = 4/10000

 ⇒ ∆R = (4/10000) x (66.67)²

 ⇒ R = 4/10000 x 4444.889

 ⇒ R = 17779.56/10000

 ⇒∴ R = 1.7779

 ⇒∴ ∆R = 1.78 Ohm

∴ total equivalent resistance in parallel combination will be :-

 ⇒ R ± ∆R

 ⇒ 66.67 ± 1.78 Ohm!!!!!!!!!!!!

$\huge\bf \red \: \boxed{\mathfrak{ \: thank \: you \: ヅ}} \:$

Answer 2

n is the number of observations, p is the number of regression parameters. 
 

Corrected Sum of Squares for Model:SSM = Σi=1n (yi^ - y)2,
    also called sum of squares for regression. 
 

Sum of Squares for Error: SSE = Σi=1n(yi - yi^)2,
    also called sum of squares for residuals. 
 

Corrected Sum of Squares Total:   SST = Σi=1n (yi - y)2
    This is the sample variance of the y-variable multiplied by n - 1.
 

For multiple regression models, we have this remarkable property: SSM + SSE = SST. 
 

Corrected Degrees of Freedom for Model:   DFM = p - 1 
 

Degrees of Freedom for Error:   DFE = n - p 
 

Corrected Degrees of Freedom Total:   DFT = n - 1
    Subtract 1 from n for the corrected degrees of freedom.
    Horizontal line regression is the null hypothesis model. 
 

For multiple regression models with intercept, DFM + DFE = DFT. 
 

Mean of Squares for Model:   MSM = SSM / DFM 
 

Mean of Squares for Error:   MSE = SSE / DFE
    The sample variance of the residuals. 
 

Mean of Squares Total:   MST = SST / DFT
    The sample variance of the y-variable. 
 

In general, a researcher wants the variation due to the model (MSM) to be large with respect to the variation due to the residuals (MSE). 
 

Note: the definitions in this section are not valid for regression through the origin models. They require the use of uncorrected sums of squares.

 

The F-test

For a multiple regression model with intercept, we want to test the following null hypothesis and alternative hypothesis:
 

H0:   β1 = β2 = ... = βp-1 = 0
 
H1:   βj ≠ 0, for at least one value of j

 
This test is known as the overall F-test for regression. 
 

Here are the five steps of the overall F-test for regression 
 

State the null and alternative hypotheses:
 

H0:   β1 = β2 = ... = βp-1 = 0 
 
H1:   βj ≠ 0, for at least one value of j

 

Compute the test statistic assuming that the null hypothesis is true:
 

F = MSM / MSE = (explained variance) / (unexplained variance)

 

Find a (1 - α)100% confidence interval I for (DFM, DFE) degrees of freedom using an F-table or statistical software. 
 

Accept the null hypothesis if F ∈ I; reject it if F ∉ I. 
 

Use statistical software to determine the p-value.


Practice Problem:  For a multiple regression model with 35 observations and 9 independent variables (10 parameters), SSE = 134 and  SSM = 289, test the null hypothesis that all of the regression parameters are zero at the 0.05 level.
 
Solution: DFE = n - p = 35 - 10 = 25 and DFM = p - 1 = 10 - 1 = 9. Here are the five steps of the test of hypothesis:
 

State the null and alternative hypothesis:
 

H0:   β1 = β2 = , ... , = βp-1 = 0 
 
H1:   βj ≠ 0 for some j


Compute the test statistic:
 

F = MSM/MSE = (SSM/DFM) / (SSE/DFE) = (289/9) / (134/25) = 32.111 / 5.360 = 5.991


Find a (1 - 0.05)×100% confidence interval for the test statistic. Look in the F-table at the 0.05 entry for 9 df in the numerator and 25 df in the denominator. This entry is 2.28, so the 95% confidence interval is [0, 2.34].  This confidence interval can also be found using the R function call qf(0.95, 9, 25).
 

Decide whether to accept or reject the null hypothesis: 5.991 ∉ [0, 2.28], so reject H0. 
 

Determine the p-value. To obtain the exact p-value, use statistical software. However, we can find a rough approximation to the p-value by examining the other entries in the F-table for (9, 25) degrees of freedom:
 
LevelConfidence IntervalF-value0.100[0, 0.900]1.890.050[0, 0.950]2.280.025[0, 0.975]2.680.010[0, 0.990]2.220.001[0, 0.999]4.71
The F-value is 5.991, so the p-value must be less than 0.005. 

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