Ma 4704 Tutorials Page

TUTORIAL SHEET 1 MA4704

TUTORIAL SHEET 1 MA4704

DiameterHeightTest

12.233.761

22.123.151

31.061.852

42.123.641

52.994.643

64.015.252

72.414.071

82.754.722

92.204.172

104.095.733

1. From the above data use the Minitab statistics package to graph the following according to the suitable data:

i. (i) Histogram(ii) Box plot (iii) Pie chart (iv) Plot Diameter Vs Height

2. Using the data from the Diameter column above calculate the following using a calculator:

i. (i) Mean(ii) Median(iii) Standard deviation(iv) inter-quartile range

b. Ans - Diameter:(i) 2.598(ii) 2.320(iii) 0.916(iv) 1.1253. Using the same data above calculate the following for each according to the suitable data using the Minitab statistics package:

i. (i) Mean(ii) Median(iii) Standard deviation(iv) inter-quartile range

b. Ans - Height:(i) 4.098(ii) 4.120(iii) 1.103(iv) 1.335

4.Give an explanation and example of the following terms:

(i) Continuous variable (ii) Discrete variable

5.An article in Quality Engineering presents viscosity data from a batch chemical process. A sample of these data are presented below (read down, then across).

13.314.9

15.8

16.0

14.513.7

13.7

14.9

15.315.2

15.1

13.6

15.314.5

13.4

15.3

14.315.3

14.1

14.3

14.815.6

14.8

15.6

15.215.8

14.3

16.1

14.513.3

14.3

13.9

14.614.1

16.4

15.2

14.115.4

16.9

14.4

14.315.2

14.2

14.0

(i) Construct a stem-and-leaf display for the viscosity data.

(ii) Construct a frequency distribution and histogram.

(iii) Convert the stem-and-leaf plot in part (i) into an ordered stem-and-leaf plot. Use this graph to assist in locating the median and the upper and lower quartiles of the viscosity data.

(iv) Calculate the following:

(a) Mean(b) mode(c) median(d) Standard deviation

(e) inter-quartile range

Ans: (a) 14.761(b) 14.3

(c) 14.7

(d) 0.846(e) 1.175

(v) Do a graphical and numerical summary of the data.

(Use Minitab to assist you in answering this question)

6. Give examples (a) Where you would use and

(b) Where you wouldnt use the following:

Bar Chart, Histogram, Pie Chart

7. The following data gives the lengths for a sample of 25 drill bits:

2529232725

2322252228

2828172430

1917232124

1520261923

(i) Calculate the standard deviation of the data above.

Ans: 3.976(ii) Summarize these data using a frequency table using 6 class intervals.

(iii) Summarize these data using a relative frequency table.

(iv) Calculate the cumulative frequency table.

(v) Draw a histogram for the data.

Formulae:Range = largest length smallest length.

Width of each interval = range / number of class intervals

(to the nearest whole number).

Frequency = proportion of observations falling into a given class.

Relative frequency expresses the frequency counts as a percentage of the total number of cases.

Spring 1998 Exam Question

8.(a) Compare and contrast the mean and the mode as appropriate measures

of central tendency (location paramatics).

(b)Describe briefly how you would construct

(i) (i) Bar chart

(ii) (ii) Histogram.

When is a bar-chart a more appropriate graphical summary than a histogram

(use data to support your answer).

(c )A machine is designed to deliver 128 mgs of powder into a container. Two

such machines are installed at the end of production lines A and B respectively.

A production manager gathered the following data from both lines.

Production Line A128,128.5,127.5,128,127.5,128,128.5,129,127.

Production Line B127,128,129.5,128.5,130,129,128,126,127.

You are required to

(i)Construct a box-plot of the data from each production line.

(ii)Use the box-plots to comment on the two machines.

(iii)The maintenance supervisor claims that because both machines place

on average the same amount of powder into each container there is no

need to overhaul either machine. Write a brief report for the operations manager in response to the maintenance supervisor

Solutions to Tutorial Sheet 1 - MA4704 will be posted here on Thursday 21st February 2002.

TUTORIAL SOLUTIONS SHEET 1 MA4704

1. Minitab Statistics Package:

(i) Histogram To draw a histogram of the Quantitative data i.e. diameter and height, you go to the GRAPH menu on the MENU BAR, then go down to histogram and click on it. From here a box appears where you enter diameter or height by double clicking on the chosen one, then click OK.

Diameter

Height

(ii)Box-plot To draw a box-plot of your quantitative data i.e. diameter and height, you go to the Graph menu on the MENU BAR, then go down to box-plot and click on it. From here a box appears where you enter diameter or height by double clicking on the chosen one, then click OK.

Diameter

Height

(iii) Pie-chart To draw a pie-chart of your qualitative data i.e. test, you go to the GRAPH menu on the MENU BAR, then go down to pie-chart and click on it. From here a box appears where you enter test in the CHART DATA IN box by double clicking on it, then click OK.

Pie-chart

(iv)Plot Diameter Vs Height - To plot diameter vs height, you go to the GRAPH menu on the MENU BAR, then go down to plot and click on it. From here a box appears where you enter diameter and height in the X and Y boxes by double clicking on them, then click OK.

Plot Diameter Vs Height

2. (i) Mean

x = data

i = 110

n = sample size

= (2.23 + 2.12 + 1.06 + 2.12 + 2.99 + 4.01 + 2.41 + 2.75 + 2.20 + 4.09)

10

= 25.98= 2.598

10

(ii) Median Calculate the median by arranging the values in increasing order. If there is an odd number of observations, the median is the middle number, find the number in the position (k + 1), where k is the number of

2

values you have. If there is an even number of observations the median is the mean of the two observations occupying the middle position, mean of the observations in position k and k +1 where k is the number of values you have.

2 2

k = 10

k = 5k +1 = 6

2 2

Diameter

11.06

22.12

32.12

42.20

52.23

62.41

72.75

82.99

94.01

104.09

Numbers in increasing order-

Median = (2.23 + 2.41) / 2 = 2.32

(iii) Standard Deviation

= mean

x = data

n = sample size

i = 1.10

s2 = (2.23 2.598) 2 + (2.12 2.598) 2 + (1.06 2.598) 2 + (2.12 2.598) 2

+ (2.99 2.598) 2 + (4.01 2.598) 2 + (2.41 2.598) 2 + (2.75 2.598) 2 + (2.20 2.598) 2 + (4.09 2.598) 2

________________

10 - 1

s2 = 7.54816 = 0.838684

9

s =

= 0.916

(iv) Inter-quartile Range To calculate the inter-quartile range we first must calculate Q1 and Q3. Calculate by arranging the values in increasing order.

Q1 is the values at position (n + 1) and Q3 is the values at position 3(n + 1).

4

4

Where n is the number of values. If the position is an integer, interpolation is used. Finally, we calculate the inter-quartile range (IQR), IQR = Q3 Q1.

Q1 = (10 + 1)= 2.75

4

Q1 is between the second and third values, three-quarters of the way up.

Q1 = 2.12 + 0.75(2.12 2.12) = 2.12

Q3 = 3(10 + 1)= 8.25

4

Q3 is between the eighth and ninth values, quarter of the way up.

Q3 = 2.99 + 0.25(4.01 2.99) = 3.245

IQR = Q3 Q1 = 3.245 2.12 = 1.125

3. Minitab Statistics Package:

Descriptive Statistics - you go to the STAT menu on the MENU BAR, then go

down to BASIC STATISTICS and then go to DISPLAY DESCRIPTIVE

STATISTICS. From here a box appears where you enter diameter or height

by double clicking on the chosen one, then click OK. Your result appear in

the session window, from which you can calculate your inter-quartile range

and draw a report from them.

- Diameter

Descriptive Statistics

Variable N Mean Median TrMean StDev

Diameter 10 2.598 2.320 2.604 0.916

Variable SE Mean Minimum Maximum Q1 Q3

Diameter 0.290 1.060 4.090 2.120 3.245

- Height



Height 10 4.098 4.120 4.175 1.103


Height 0.349 1.850 5.730 3.517 4.85

4.Quantitative data can be quantified i.e. they are associated with specific numerical values.

Qualitative data can only be separated into different categories.

(i) Continuous Variable

Quantitative data, random variables that can assume a value corresponding to any point on some interval, without gaps.

Example: length of a drill bit

(ii) Discrete Variable

Quantitative data, usually associated with data that involves counting, may have gaps.

Example: Number of defects in a manufacturing process.

5. (i) Stem and Leaf Display

Using the Minitab Statistics Package, you go to the GRAPH menu on the

MENU BAR, then go down to stem-and leaf. From here a box appears

where you enter the column your data is in by double clicking on the

chosen one, then click OK.

Character Stem-and-Leaf Display

Stem-and-leaf of C5 N = 44

Leaf Unit = 0.10

3 13 334

7 13 6779

18 14 01112333334

(8) 14 55568899

18 15 1222233334

8 15 6688

4 16 014

1 16 9

(ii) Frequency distribution

Using the Minitab Statistics Package, you go to the STAT menu on the

MENU BAR, then go down to TABLES and then TALLY. From here a

box appears where you enter the column your data is in by double clicking

on the chosen one, then click on the box with PERCENTS on it to get

percents and then click OK.

Summary Statistics for Discrete Variables

C5 Count Percent

13.3 2 4.55

13.4 1 2.27

13.6 1 2.27

13.7 2 4.55

13.9 1 2.27

14.0 1 2.27

14.1 3 6.82

14.2 1 2.27

14.3 5 11.36

14.4 1 2.27

14.5 3 6.82

14.6 1 2.27

14.8 2 4.55

14.9 2 4.55

15.1 1 2.27

15.2 4 9.09

15.3 4 9.09

15.4 1 2.27

15.6 2 4.55

15.8 2 4.55

16.0 1 2.27

16.1 1 2.27

16.4 1 2.27

16.9 1 2.27

N= 44

Histogram

To draw a histogram of the Quantitative data, you go to the GRAPH menu on the MENU BAR, then go down to histogram and click on it. From here a box appears where you enter whichever column your data is in by double clicking on the chosen one, then click OK.

(iii) Ordered Stem-and-Leaf Plot

- arranges the leaves by magnitude, which is produced automatically by Minitab.

(iv) Descriptive Statistics - you go to the STAT menu on the MENU BAR,

then go down to BASIC STATISTICS and then go to DISPLAY

DESCRIPTIVE STATISTICS. From here a box appears where you enter

the column in which your data is contained by double clicking on the

chosen one, then click OK. Your result appear in the session window,

from which you can calculate your inter-quartile range and draw a report

from them.



C5 44 14.761 14.700 14.740 0.846


C5 0.127 13.300 16.900 14.125 15.300

You can also calculate these using a calculator.

_

(a) Mean

x = data

i = 110

n = sample size

= 649.5 = 14.761

44

(b) Mode of a set of values is the value that occurs most frequently.

We can calculate this by counting the number of times a value appears or

by using the Minitab Statistics Package, you go to the STAT menu on the

MENU BAR, then go down to TABLES and then TALLY. From here a

box appears where you enter the column your data is in by double clicking

on the chosen one, then click OK. This gives you a frequency distribution.

Mode = 14.3 which occurs 5 times.

(c) Median Calculate the median by arranging the values in increasing order. If there is an odd number of observations, the median is the middle number, find the number in the position (k + 1), where k is the

2

number of values you have. If there is an even number of observations the median is the mean of the two observations occupying the middle position, mean of the observations in position k and k +1 where k is the

2 2

number of values you have.

k = 44

k = 22

k +1 = 23

2

2

Numbers in increasing order-

113.31214.22314.83415.3

213.31314.32414.83515.3

313.41414.32514.93615.4

413.61514.32614.93715.6

513.71614.32715.13815.6

613.71714.32815.23915.8

713.91814.42915.24015.8

8141914.53015.24116

914.12014.53115.24216.1

1014.12114.53215.34316.4

1114.12214.63315.34416.9

Median = (14.6 + 14.8) / 2 = 14.7

(d)Standard Deviation

= mean

x = data

n = sample size

i = 1.10

s2 = 30.91

44 - 1

s2 = 0.718837209

s =

= 0.846

(e) Inter-quartile Range To calculate the inter-quartile range we first must calculate Q1 and Q3. Calculate by arranging the values in increasing order.

Q1 is the values at position (n + 1)

4

Q3 is the values at position 3(n + 1).

4

Where n is the number of values. If the position is an integer, interpolation is used. Finally, we calculate the inter-quartile range (IQR),

IQR = Q3 Q1.

Q1 = (44 + 1)= 11.25

4

Q1 is between the eleventh and twelfth values, quarter of the way up.

Q1 = 14.1 + 0.25(14.2 14.1) = 14.125

Q3 = 3(44 + 1)= 33.75

4

Q3 is between the thirty-third and thirty-fourth values, three-quarters of the way up.

Q3 = 15.3 + 0.75(15.3 15.3) = 15.3

IQR = Q3 Q1 = 15.3 14.125 = 1.175

(v) Graphical and Numerical Summary

N

There are 44 samples in the data set.

Mean The mean diameter is 14.761.

MedianThe median diameter is 14.7. The mean is larger then the median indicating the distribution may be skewed to the right.

Tr MeanThis is about 90% trimmed mean. It is the average of the data values after removing the smallest 5% (rounded to the nearest integer) and the largest 5%. Since 5% of 44 is 2.2 or 2, the smallest 2 and the largest 2 values are removed and the mean of the remaining 40 values are calculated. The trimmed mean of 14.740 is slightly more than the untrimmed.

StDev The sample standard deviation is 0.846

SE MeanThe standard deviation of the mean is the sample standard deviation (0.846) divided by the square root of the sample size (44). We use this measure to make inferences.

Min

The minimum diameter is 13.3.

Max

The maximum diameter is 16.9.

Q1

About 25% of the diameters are less than 14.125.

Q2About 75% of the diameters are less then 15.3 and about 25% of the diameters are greater than 14.125.

About 50% of the batch, have viscosity between 14.125 and 15.3; the median is about 14.7. The area at either side of the median seem to be divided evenly, the whisker to the right hand side is longer than the whisker to the left hand side indicating the distribution is skewed towards the higher viscosity. The box-plot did not detect any outliers.

The histogram groups viscosity into classes of width 0.5. The first class is centered on 13.5 and runs from 13.3 to less than 13.7. the second class is centered at 14.0 and runs from 13.7 to less than 14.3, and so on. The distribution seems to be centered at about 15.0 and is slightly skewed toward higher viscosity.

Character Stem-and-Leaf Display

Stem-and-leaf of C5 N = 44

Leaf Unit = 0.10

3 13 334

7 13 6779

18 14 01112333334

(8) 14 55568899

18 15 1222233334

8 15 6688

4 16 014

1 16 9

The N = 44 indicates that 44 values are in the stem and leaf display. The leaf unit stated in the display is 0.10; which means the stem unit is 1. The smallest value, with a stem of 13 and a leaf of 4, have a viscosity of 13.4; 3 batches have a viscosity if 13.4. The largest value have a viscosity of 16.9. The center is at about 14.9.

Conclusion:

From the information gathered and analysis we conclude that the batch from the chemical process seems to show that most of the process is producing batches with high viscosity.

6.

(a) Would Use(b) Wouldnt Use

Bar ChartQualitative Data:

Test results in a factoryLength of drill bits

HistogramQuantitative Data:

Length of drill bitsTest results in a factory

Pie ChartQualitative Data:

Test results in a factoryLengths of drill bits

7.(i) Standard Deviation = mean

x = data

n = sample size

i = 1.10

Mean =

s = =

(ii)/(iii)/(iv) Range = largest length smallest length.

= 30 15 = 15 Width of each interval = range / number of class intervals

Class IntervalFrequencyRelative FrequencyCumulative Rel. Freq

14.5 17.531212

17.5 20.531224

20.5 23.572852

23.5 26.562476

26.5 29.552096

29.5 32.514100

Totaln = 25100%

= 15 / 6 = 3 (to the nearest whole number).

To calculate the cumulative relative frequency, the first figure is the first figure in the relative frequency column and then for the second figure, we add the first figure in the relative frequency to the second, and then for the third we add this number to third value in the relative frequency column and so on and so on.

(v)Histogram To draw a histogram of the Quantitative data i.e. lengths , you go to the GRAPH menu on the MENU BAR, then go down to histogram and click on it. From here a box appears where you enter the column containing your data by double clicking on the chosen one, then click OK.

- histogram of the data:

Exam Question- Spring 1998

8. (a) The mean or average is the sum of the observations divided by the total number of observations. It is used when we get a bell-shaped curve for the data. It is the central tendency in the distribution, or its location.

The mode of a set of data is defined as the number which occurs most frequently.

The mean is not necessarily the fiftieth percentile of the distribution (the median), and it is not necessarily the most likely value of the variable ( which is called the mode). The mean simply determines the location of the distribution.

We usually use the mean or the median to describe the center of our distribution. The mean is sensitive to extreme values whereas the median is unaffected by extreme values.

If the histogram is symmetric this implies that the mean is equal to the median which is equal to the mode. The most commonly used measure of centrality is for a symmetric distribution is the mean.

If the histogram is asymmetric i.e. skewed this implies we will use the median as our measure of centrality since it will be unaffected by the values that are causing the histogram to be skewed.

When we have a left-skewed distribution the mean is less than the median. When we have a right-skewed distribution the mean is greater than the median.

(b) Constructing:

(i) Bar Chart

Draw a vertical axis about two and a half centimeters or one inch in from the left-hand side of the graph paper and a horizontal one a similar distance from the foot of the paper. These axes will intersect. The point of intersection on the vertical axis is zero. The horizontal one may be started at any convenient point near the vertical axis.

Choose scales which will show your diagram to the best advantage, and will include all your data.

Label your axes with appropriate title and mark in numerical scale values.

All bars in a bar diagram should be of the same width.

(ii) Histogram

To draw a histogram, we must first group our data into intervals. Then we place a number line on the horizontal axis. The number line should start at the smallest value in the first interval and continue until the largest value in the last interval. On the vertical scale, we mark off count values or proportion values starting with zero.

Then for each interval, we make a bar with the width covering the interval and height equal to the corresponding count or proportion.

A bar chart is a more appropriate graphical summary then a histogram when we have categorical data such as car manufacturers. This implies we can list the different brands along the horizontal axis.

(c) (i)Box-plot -

We need to calculate the median and upper and lower quartiles

before we start to draw a box-plot. Then, draw a box between the

upper and lower quartiles, thus indicating where the middle 50%

of the data fall. Horizontal lines called whiskers are then extended

from the middle of the sides of the box to the minimum and to the

maximum. The median is then marked with a vertical line inside

the box.

(ii)For Line A about 50% of the production is between 127.5 and 128.5, for Line B about 50% of the production is between 127 and129.25, both line productions have a median of 128. Production Line A, appears to run more smoothly then Production Line B as A has equal distance at both sides of the median unlike B where there is slightly more area above the median then below. In Production Line A and B, the whiskers appear to have the same length which implies both distributions are not skewed. The whiskers only extend to the last data value within 1.5 IQR, and anything outside this is considered an outlier. Outliers are marked individually, sometimes by an o. Both box plots did not detect any outliers.

(iii) Line A appears to have a mean of 128 which implies it hits its target delivery the majority of the time. Line B, however has a mean of 128.11 which is slightly over the target delivery and so in bulk production would lose a lot of its power by overfilling the containers. Therefore, I would recommend that Production Line B needs to be overhauled.

TUTORIAL SHEET 2 MA47041. The following data are the joint temperatures of the O-rings (degrees F) for each test firing or actual launch of the space shuttle rocket motor:

8449614083674566

7069805868606772

7370576370785267

5367756170817679

75765831

(a) Compute the sample mean and sample standard deviation.

Ans:Mean 65.86Standard Deviation 12.16

(b) Find the median and the upper and lower quartiles.

Ans:Median 67.50

Upper Quartile 75.00

Lower Quartile 58.50

(c) Set aside the smallest observation (31oF) and recomputed the quantities in parts (a) and (b).

Ans:Mean 66.86Standard Deviation 10.74Median 68.00

Upper Quartile 75.00

Lower Quartile 60.00

Comment on your findings. How different are the other temperatures from this smallest value.

2. For an electronic manufacturing process, it currently has specifications of 100 10 milliamperes. The process mean ( and standard deviation ( are 107.0 and 1.5 respectively. Calculate the process capability and interpret this value when the process demands a minimum Cp of 2.0. Will the product be accepted?

Ans: Cp = 2.223.An engineering company you supply demands a minimum Cp of 1.33. One component they require, has a target length of 80mm. The product specification defines an acceptable error of +/- 5mm. The process standard deviation is ( = 1.5. Will the product be accepted, give a reason for your answer.

Ans: Cp = 1.111

4.Give a definition of the following: (a) Signal (b) Noise.

5.A manufacturer of coil springs produces a standard length to be 10cm. It is agreed that a 5% probability of a Type 1 Error (signal : noise = +/- 1.96) is acceptable. The standard deviations is 1.86cm. A sample of 6 is taken from the production line and the following measurements were taken:

x1 = 9

x2= 12

x3=7

x4=9

x5=8

x6=11

Calculate the signal to noise ratio and give a conclusion to your findings.

Ans:Signal : Noise Ratio = 0.8784

6. (a) Assume that Z has a standard normal distribution. Use the tables to determine the

value for z that solves the following:

(i) P(-z < Z < z) = 0.95

Ans:z = +/- 1.96(ii) P(-z < Z < z) = 0.99

Ans:z = +/- 2.575(iii) P(-z < Z < z) = 0.9973

Ans:z = +/- 3

(b) Assume that X is normally distributed with a mean of 6 and a standard deviation of 3.

Determine the value for x that solves each of the following:

(i) P(X > x) = 0.5

Ans:x = 7.5(ii) P(X > x) = 0.95

Ans:x = 8.85(iii) P(x < X < 9) = 0.2

Ans:x = 7.08(iv) P(3 < X < x) = 0.8

Ans:x = 7.97.The length of an injected-molded plastic case that holds magnetic tape is normally

distributed with a mean length of 90.2 millimeters and a standard deviation of 0.1 millimeter.

(a) What is the probability that a part is longer then 90.3 millimeters or shorter than 89.7 millimeters?

Ans: 0.1587(b) What should the process mean be set at to obtain the greatest number of parts between 89.7 and 90.3 millimeters?

Ans: 90(c) If the parts are not between 89.7 and 90.3 millimeters are scrapped, what is the yield for the process mean that you selected in part (b)?

Ans: 0.9973Spring 1999 Exam Question

8.(a)Assume that the length of catheter manufactured for the Health Care

industry are normally distributed with a mean of 100 mm and a standard

deviation of 1mm. Draw a rough sketch and then calculate the

corresponding probabilities for the following measurements occurring on an

individual catheter

(i)Between 100 and 101.5 mms.

Ans: 0.4332

(ii)Less than 98.6 mm

Ans: 0.0808

(iii)Between 98.5 and 101 mm

Ans: 0.7745

(iv)Less than 100.5 mms.

Ans: 0.6915

(b)Assume that z scores are normally distributed with a mean of zero and a

standard deviation of 1.

(i)If p(0 < z < a)

= 0.388find a

Ans: a = 1.215

If p(-b < z < -b)= 0.7698 find b

Ans: b = 0.295

If p(z < c)

= 0.1841find c

Ans: c = 0.9

(c)The mean time between failures for prototypes of computer printers is

known to be approximately normally distributed with mean

and variance

. A project engineer has found that 80% of the printer prototypes last

for at least 24 hours continuous use and that 6% last for 44 or more hours

continuous use. Calculate the mean time between failures and the standard

deviation based on the project engineers results.

(d)The electronic components of the printer follow an exponential distribution

with a mean life of 200 hours. What is the probability that the electronics will

(a)fail in the first twenty four hours.

(b)Survive for more than 44 hours.

Solutions to Tutorial Sheet 2 - MA4704 will be posted here on Thursday 28th February 2002.

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