Steve Goddard

Contents

Topic Page

Presentation Of Data 1Measures of Central Tendency and Dispersion 3Linear Correlation and Linear Regression 4Probability, Binomial and Poisson Distribution, Sampling and Estimation Theories and Significance Testing

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Bibliography 11

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Steve Goddard

Analytical Methods – Assignment 4

Statistics and Probability

Presentation of Data

The diameter in millimeters of a reel of wire is measured in 48 places and the results are as shown:

2.10 2.29 2.32 2.21 2.14 2.222.28 2.18 2.17 2.20 2.23 2.132.26 2.10 2.21 2.17 2.28 2.152.16 2.25 2.23 2.11 2.27 2.342.24 2.05 2.29 2.18 2.24 2.162.15 2.22 2.14 2.27 2.09 2.212.11 2.17 2.22 2.19 2.12 2.202.23 2.07 2.13 2.26 2.16 2.12

1.1 Form a frequency distribution of the diameters having about 6 classes.

RangeFrequency

2.05 ≥ x ≤ 2.09 32.10 ≥ x ≤ 2.14 112.15 ≥ x ≤ 2.19 112.20 ≥ x ≤ 2.24 122.25 ≥ x ≤ 2.29 92.30 ≥ x ≤ 2.34 2

1.2 Draw a histogram depicting the data

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1.3 Form a cumulative frequency distribution

Range Frequency Cumulative Frequency2.05 ≥ x ≤ 2.09 3 32.10 ≥ x ≤ 2.14 11 142.15 ≥ x ≤ 2.19 11 252.20 ≥ x ≤ 2.24 12 372.25 ≥ x ≤ 2.29 9 462.30 ≥ x ≤ 2.34 2 48

1.4 Draw an Ogive for the data

I went by the point made by a website I found about Ogives and drew straight lines rather than a smooth one.

“The points should be connected by straight lines, not a smooth curve as taught by many textbooks.  The straight line assumes that the data are uniformly distributed across the class interval - as represented on a histogram by a rectangular bar.  Whilst this will almost certainly not be true, it gives an objective view that allows different people to come to common conclusions.   Connecting the points by a smooth curve is subjective and will lead to different people drawing different conclusions from the data.”

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Measures of Central Tendency and Dispersion

2. Determine the mean, median and mode for the following data set:

73.8, 126.4, 40.7, 141.7, 28.5, 237.4, 157.9

Mean

Median = 126.4

Mode = No mode in the set

3. The values of capacitances, in microfarads, of ten capacitors selected at random from a large batch of similar capacitors are:

34.3, 25.0, 30.4, 34.6, 29.6, 28.7, 33.4, 32.7, 29.0, 31.3

Determine the standard deviation from the mean for these capacitors, correct to three significant figures.

First of all I worked out the statistical average:

Next I had to work out the deviation each result was from the statistical average:

Capacitance Value

Deviation Deviation2

34.3 3.4 11.5625.0 5.9 34.8130.4 0.5 0.2534.6 3.7 13.6929.6 1.3 1.6928.7 2.2 4.8433.4 2.5 6.2532.7 1.8 3.2429 1.9 3.6131.3 0.4 0.16

Total 80.1

I then used the formulae: Standard Deviation

Putting this into context with the values I have been given gives,

μF

Linear Correlation and Linear Regression

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Steve Goddard

4. In an experiment to determine the relationship between the current flowing in an electrical circuit and the applied voltage, the results obtained are:

Current (mA) 5 11 15 19 24 28 33Applied voltage (V) 2 4 6 8 10 12 14

Determine, using the product-moment formulae, the coefficient of correlation.

First of all I worked out the mean of x and y, x being current and y being applied voltage.

Mean value of x = 19.285Mean value of y = 8

Next I calculated the deviation of each result:

X Deviation x = Y Deviation y =

5 -14.285 2 -611 -8.285 4 -415 -4.285 6 -219 -0.285 8 024 4.715 10 228 8.715 12 433 13.715 14 6

Knowing the product-moment formulae for working out the coefficient for linear correlation is given by:

I then used the following table to get the required results to put into the equation

xy X2 Y2

85.71 204.061 3633.14 68.641 168.57 18.361 40 0.081 09.43 22.231 434.86 75.951 1682.29 188.10 36

254 = 577.426 = 112

Putting these values into the equations give:

This means the coefficient of linear correlation = 0.9987

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Steve Goddard

5. The relationship between the voltage applied to an electrical circuit and the current flowing is as shown:

Current (mA) Applied Voltage (V)2 54 116 158 1910 2412 2814 33

Assuming a linear relationship, determine the equation of the regression line of applied voltage, Y, on current, X, correct to 4 significant figures.

This table will show the results I need to put into the equation,

Current (mA), X

Applied Voltage (V), Y

X2 XY Y2

2468

101214

5111519242833

4163664

100144196

104490

152240336462

25121225361576784

1089

56 135 560 1334

3181

The regression coefficients a0 and a1 can be obtained using the equations:

And

Where N is the number of co-ordinate values.

Substituting in the results from the above table gives:

I will call the top equation 1 and the bottom 2.

If I multiply each part of equation 1 by 56 and each part of the bottom equation by 7 I will get:

I will call the top equation 3 and the bottom 4.If I subtract equation 3 from equation 4 it will give me:

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Steve Goddard

So rearranging the equation to make a1 the subject you would get:

Substituting this answer into equation 1 allows me to work out the value for a0 as follows:

135 = 7 a0 + 56 (2.2679)

Thus putting a0 and a1 into the first equation will give:

The equation of the regression line of applied voltage on current =

I checked this result out in excel by drawing the graph and making a linear trend line. From this I checked to see if my results look similar to the graph, which they did.

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0

5

10

15

20

25

30

35

0 2 4 6 8 10 12 14 16

Series1

Linear (Series1)

Steve Goddard

Probability, Binomial and Poisson Distributions, Sampling and Estimation Theories and Significance Testing

6. A box of fuses are all of the same shape and size and comprises 23 off 2A fuses, 47 off 5A fuses and 69 off 13A fuses. Determine the probability of selecting at random, a:

2A fuse

5A fuse

13A fuse

Probability of getting a 2A fuse =

Probability of getting a 5A fuse =

Probability of getting a 13A fuse =

7. An automatic machine produces, on average, 10% of its components outside of the tolerance required. In a sample of 10 components from this machine, determine the probability of having three components outside of the tolerance required by assuming binomial distribution.

I did this by using the formulae:

When p = probability of component being outside tolerance and q being a component inside of tolerance. q + p should always add up to be 100% or 1 Also n is the number of number of components used.

Next I worked out the probability of 3 components being defective:

I used the formula showing the binomial expansion of :

….. And so on.

This enabled me to work out the equations for:

No defective components:

1 defective component:

2 defective components:

And finally

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Steve Goddard

3 defective components

This makes the probability of selecting 3 incorrect components from a batch of 10: 5.73%

To check my answers I used an online binomial calculator (see below)

I typed the required data in and hit generate and it came up with the possibilities from 0 -10.

I was satisfied with my results because they matched the table generated by the calculator

8. A large batch of electric light bulbs has a mean time to failure of 800 hours and the standard deviation of the batch is 60 hours. Using a table

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Steve Goddard

of partial areas under the standardized normal curve, determine the probability that the mean time to failure will be less than 785 hours (correct to three decimal places), if a random sample of 64 light bulbs is drawn from the batch.

Mean time to fail = 800hoursStandard deviation = 60hours

Find probability that the mean is under 785hours,

First of all I will work out the standard areas of the means which is given by the equation:

Where is the standard deviation and N is the sample size

Deviations

Once I had found the value for z I looked at the table that stated the partial areas under the standardised curve. From this I could see the z value of -2 equates to a partial area of -0.4772.

Next I have to subtract the -0.4772 from 0.5000 because this is the value where z = 0.The probability of the mean time to failure being less than 785 hours is 0.023 to 3dp.

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Steve Goddard

9. The specific resistance of a reel of German silver wire of nominal diameter 0/5mm is estimated by determining the resistance of 7 samples of the wire. These were found to have resistance values, in ohms per meters, as follows:

1.12, 1.15, 1.10, 1.14, 1.15, 1.10, 1.11

Determine the 99% confidence interval for the true specific resistance of the reel of wire.

x Deviation Deviation2 1.12 -0.004 0.0000161.15 0.026 0.0006761.10 -0.024 0.0005761.14 0.016 0.0002561.15 0.026 0.0006761.10 -0.024 0.0005761.11 -0.014 0.000196

7.87 0.002972

Firstly to work out the standard deviation I used the normal formula:

To find the confidence interval for the specific resistance use I used the equation:

Using the table in my notes I managed to work out that:

For a 99% confidence level = a confidence coefficient of 2.58

N = number of samples

So:

This means that the 99% confidence interval for the true specific resistance of the reel of wire is from 1.15 to 1.10 ohms.

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Steve Goddard

Bibliography

www.efunda.com -– Engineering Resource Website

www. Wi kipedia.com – Online Encyclopedia

www.google.com – Search Engine

http://onlinestatbook.com/chapter5/binomial.html - Binomial Distribution Notes

http://www.adsciengineering.com/bpdcalc/ - Binomial Distribution Calculator

NC Statistics notes

HNC Statistic Notes

Engineering Mathematics by K A Stroud

http://home.ched.coventry.ac.uk/Volume/vol0/ogive.htm - Explanation of an Ogive

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