Question

ttps://mail yahoo com/d/search/key ord-numerical%25 20analysis%2520test/messages AP4VBdkk6Bai JosPgHwGA6DZ8g/AP4VBdkk6Ba 1 sA 6 created and used should be submitted as email attachments (labled accordigly) be fore exitingthe ezamination. 1. The acceleration due to gravity (9, m/s2) at an altitude (h, Km) above the surface of the earth is given by the following experimental data set 9 9.81 9.78 9.76 9.74 9.72 9.68 9.63 9.61 9.59 9.52 Find and plot the linear least squares (regression) fit of a straight line to the given data and use this modelto estimate gat h 67 Km. Does this linear regression model have"merit'? Justify your answer by comparing the standard error of the estimate to the standard deviation. 2. Consider the experimental data, from a wind tunnel experiment, given in the table below; where vis velocity(m/s, independent variable) and F is force (N, dependent variable). 10 20 30 40 50 6 7 80 9100 115 120 F 32 65 270 467 580 1150 790 1460 750 1360 1470 1490 U se this data in conjunction with the following methods to estimate Fat v 82 m/s: (a) Ezact polymomial fit (b) Least squares linear regression (c) Newton's interpolating polmomial (a) Lagrange interpolating polymomial

Answer 1

Given:

10 | 20 | 25 | 32 | 40 | 55 | 60 | 63 | 75 g 9.81 9.78 9.76 9.74 9.72 9.68 9.63 9.61 9.59 9.52

Lets interpolate the points

(32,9.72) (63,9.59)( (75,9.52) (0,9.81) 20 40 60 80

Since the graph is linear ,we can find the value of g at h = 67(by simple interpolation method)

63	9.59
75	9.52

75-63 = 12 so at 12 kn altitude gravity difference = 9.59-9.52 = 0.07

at 67km the difference ==> 67-63 = 4km

==> 0.07/12 *4 = 0.0233

So at 67km the gravity difference = 9.59-0.0233 = 9.5667

lets calculate the standard deviation:

Sample mean : $\mu _x = \sum x / n$

$\mu _x = \frac{9.81+9.78+9.769.74+9.72+9.68+9.63+9.61+9.59+9.52}{10}$

$\mu _x = 9.684$

standard deviation :

$\sigma _x=\sqrt{\sum_{i=1}^{n}\frac{(x_i-\mu _x)^2}{n-1}}$

$\sigma _x=0.0931$

Standard error:  $\frac{ \sigma _x}{ \sqrt{n}} = \frac{0.0931}{\sqrt{10}} = 0.0294$

The linear regression model have merit because $\sigma _x=0.0931$ > 0.0233

means deviated in interpolation is lower and deviation allowed is more.So the calculation that we did by graph have the merit and error is 2.9% only but allowed wrror is 5%.So interpolation method used have significance or have merit regards those two standard deviation and standard error.