Question

5. In lab #2 we used LINEST (Excel function) to determine the slope and uncertainties of the slope. Imagine that you are performing the following measurements: X Y 1 2.2 2 3.1 3 4.5 4 6.3 5 7.5 6 8.3 7 9.7 8 10.5 9 11.2 10 12.5

5.a. Using LINEST, calculate the slope and uncertainty of the slope for given data. Hint: Set your y column as your dependent variable and x column as your independent variable.

5.b. Calculate the mean value x , the standard deviation V y ,and standard deviation of the mean σ y of your ‘measured’ quantity X over N trials (number of runs). Xi (units) 0 .65 0.68 0.62 0.7 0.67 0.63 0.61 0.6 0.66 0.64

PLEASE SHOW WORK

Answer 1

5. Imagine that we are performing the following measurements which given below as -

X : 1   2      3     4    5     6     7      8       9     10   

Y : 2.2 3.1 4.5 6.3 7.5 8.3 9.7 10.5 11.2 12.5

5. (a) Using linest, the slope and uncertainty of the slope for given data will be given as :

slope means "mean value" and "Uncertainity means "standard deviation"

mean value, X_mean = (2.2 + 3.1 + 4.5 + 6.3 + 7.5 + 8.3 + 9.7 + 10.5 + 11.2 + 12.5) / 10

X_mean = 75.8 / 10

X_mean = 7.58

and standard deviation is given by an equation,    = { [ (X_mean - X_i)² / (n - 1)]^1/2    

where, (X_mean - X₁)² = (7.58 - 2.2)² = 28.94

(7.58 - 3.1)² = 20.07

(7.58 - 4.5)² = 9.48

(7.58 - 6.3)² = 1.63

(7.58 - 7.5)² = 0.0064

(7.58 - 8.3)² = 0.5184

(7.58 - 9.7)² = 4.49

(7.58 - 10.5)² = 8.52

(7.58 - 11.2)² = 13.10

(7.58 - 12.5)² = 24.20

sum of all values => 110.95

= [(110.95) / (10 - 1)]^1/2

= (12.32)^1/2

= 3.509

(slope uncertainity) = (7.58 3.509)

5. (b) X_i (units) : 0.65 0.68 0.62 0.7 0.67 0.63 0.61 0.6 0.66 0.64

mean value, X_mean = sum of all measured quantity / total number of trials                                              { eq.1 }

X_mean = (0.65 + 0.68 + 0.62 + 0.7 + 0.67 + 0.63 + 0.61 + 0.6 + 0.66 + 0.64) / 10

X_mean = 6.46 / 10

X_mean = 0.646

standard deviation is given by an equation,    = { [ (X_mean - X_i)² / (n - 1)]^1/2                                         { eq.2 }

where, (X_mean - X₁)² = (0.646 - 0.65)² = 1.6 x 10^-5

(0.646 - 0.68)² = 0.001156

(0.646 - 0.62)² = 0.000676

(0.646 - 0.7)² = 0.002916

(0.646 - 0.67)² = 0.000576

(0.646 - 0.63)² = 0.000256

(0.646 - 0.61)² = 0.001296

(0.646 - 0.6)² = 0.002116

(0.646 - 0.66)² = 0.000196

(0.646 - 0.64)² = 3.6 x 10^-5

=> (1.6 x 10^-5 + 0.001156 + 0.000676 + 0.002916 + 0.000576 + 0.000256 + 0.001296 + 0.002116 + 0.000196 + 3.6 x 10^-5)

=> (5.2 x 10^-5 + 0.009188)

=> 0.00924

= [(0.00924) / (10 - 1)]^1/2

= (0.001026667)^1/2

= 0.03204

standard deviation of the mean is given by :

_mean = / n                                                                      { eq.3 }

inserting the values in eq.3,

_mean = (0.03204) / 10

_mean = (0.03204) / 3.162

_mean = 0.01013