Statistics and Probability Assignment on Hypothesis Testing January 30, 2023 Instructions This document contains the questions for your assignment project on Statistical Testing. The questions refer to the data given in the individual worksheets in Excel document ‘Assignment Datasets.xlsx’. Please read the following points. 1. All submissions must be in the form of PDF documents. Spread- sheets exported to PDF will be accepted, but calculations must be annotated or explained. 2. It is up to you how you do the calculations in each question, but you must explain how you arrived at your answer for any given calculation. This can be done with a written explanation and by using the relevant equations, along with showing the results of intermediate stages of the calculations. In other words, you need to show that you know how to do a calculation for a statistic other than using spreadsheet functions. 3. Each one of the questions involves a statistical test. Marks within each question will generally be awarded for: 1 • Deciding which statistical test to use, • Framing your Hypotheses and proper conclusions, • Identifying the parameters for the test and • Showing a reasonable level of clarity, detail and explanation in the calculations needed to carry out the test. 4. The data you have been given is in the worksheets of an Excel spreadsheet. This spreadsheet is locked against editing. Please to not try to circumvent this; if you wish to use a spreadsheet to do your calculations, you should copy and paste your data into your own spreadsheet and work with that. Question 1 The lifetimes (in units of 106 seconds) of certain satellite components are shown in the frequency distribution given in ‘Dataset1’. 1. Draw a frequency polygon, histogram and cumulative frequency polygon for the data. 2. Calculate the frequency mean, the frequency standard deviation, the median and the first and third quartiles for this grouped data. 3. Compare the median and the mean and state what this indicates about the distribution. Comment on how the answer to this ques- tion relates to your frequency polygon and histogram. 4. Explain the logic behind the equations for the mean and standard deviation for grouped data, starting from the original equations for a simple list of data values. (This does not just mean ’explain how the equations are used’.) Page 2 5. Carry out an appropriate statistical test to determine whether the data is normally distributed. Question 2 A manufacturer of metal plates makes two claims concerning the thickness of the plates they produce. They are stated here: • Statement A: The mean is 200mm • Statement B: The variance is 1.5mm2. To investigate Statement A, the thickness of a sample of metal plates produced in a given shift was measured. The values found are listed in Part (a) of worksheet ‘Dataset2’, with millimetres (mm) as unit. 1. Calculate the sample mean and sample standard deviation for the data in Part (a) of ’Dataset2’. Explain why we are using the phrase ’sample’ mean or sample’ standard deviation. 2. Set up the framework of an appropriate statistical test on State- ment A. Explain how knowing the sample mean before carrying out the test will influence the structure of your test. 3. Carry out the statistical test and state your conclusions. To investigate the second claim, the thickness of a second sample of metal sheets was measured. The values found are listed in Part (b) of worksheet ‘Dataset2’, with millimetres (mm) as unit. 1. Calculate the sample mean and then the sample variance and standard deviation for the data in Part (b). Page 3 2. Set up the framework of an appropriate statistical test on State- ment B. Explain how knowing the sample variance before carry- ing out the test would influence the structure of your test. 3. Carry out the statistical test and state your conclusions. Question 3 A manager of an inter-county hurling team is concerned that his team lose matches because they ‘fade away’ in the last ten minutes. He has measured GPS data showing how much ground particular players cover within a given time period; this is the data in list (a) in worksheet ‘Dataset3’. He has acquired the corresponding data from an opposing, more successful team, which is given in list (b). 1. Calculate the sample mean and sample standard deviation for the two sets of data. 2. Set up the frame work of an appropriate statistical test to deter- mine whether there is a difference in the distances covered by the two groups of players. 3. Explain how having the results of the calculations above in ad- vance of doing your statistical test will influence the structure of that test. 4. Carry out the statistical test and state your conclusions. Question 4 A study was carried out to determine whether the resistance of the control circuits in a machine are lower when the machine motor is Page 4 running. To investigate this question, a set of the control circuits was tested as follows. Their resistance was measured while the machine motor was not running for a certain period of time and then again while the motor was running. The values found are listed in worksheet ‘Dataset4’, with kilo-Ohms as the unit of measurement. 1. Set up the structure of an appropriate statistical test to determine whether the resistance of the control circuit in a machine are lower when the machine motor is running. 2. Explain how the order of subtraction chosen to calculate the dif- ferences will influence the structure of the test. 3. Give a reason why the data is measured with the engine not run- ning first and then with the engine running. 4. Explain how knowing the mean of the differences in advance will influence the structure of your statistical test. 5. Carry out the statistical test and state your conclusions. Question 5 A study was carried out to determine the influence of a trace element found in soil on the yield of potato plants grown in that soil, defined as the weight of potatoes produced at the end of the season. A large field was divided up into 14 smaller sections for this experiment. For each section, the experimenter recorded the amount of the trace element found (in milligrams per metre squared) and the corresponding weight of the potatoes produced (in kilograms). This information is presented in the worksheet ‘Dataset5’ in the Excel document. Define X as the trace element amount and Y as the yield. Page 5 1. Draw a scatterplot of your data set. 2. Calculate the coefficients of a linear equation to predict the yield Y as a function of X. 3. Calculate the correlation coefficient for the paired data values. 4. Set up the framework for an appropriate statistical test to estab- lish if there is a correlation between the amount of the trace ele- ment and the yield. Explain how having the scatterplot referred to above and having the value of r in advance will influence the structure of your statistical test. 5. Carry out and state the conclusion of your test on the correlation. 6. Comment on how well the regression equation will perform based on the results above. Question 6 A multinational corporation is conducting a study to see how its em- ployees in five different countries respond to three gifts in an incentive scheme. The numbers of employees who choose each of the three gifts (G1 to G3) in each of the five countries (A to E) are given in the table in ‘Dataset6’ in the Excel document. 1. Set up the structure of an appropriate statistical test to deter- mine whether the data supports a link between choice of gift and country, including the statistic to be used. 2. Carry out this test, showing clearly in your work how the expected values are calculated for your test statistic. Page 6 Dataset 1 Assignment : Hypothesis Testing Type the last three digits of your student number in the green cell: 678 -7.9549632153 14.1984112513328 Dataset 1 7 20GroupsFrequencies 0.1300to307120.2768276259 0.8307to314184.4940491225 620.67314to3214428.5400756165 0.80.76321to3288871.6823966744 1328to3358671.6823966744 0.73335to3424128.5400756165 6.95461382910.45342to349154.4940491225 8.84555462860.42349to35690.2768276259 14.19841125130.14 8.26373677340 2.0212369411 1.2742778891 -4.9179951627 -7.9549632153 &"Helvetica Neue,Regular"&12&K000000&P Dataset 2 -37.4147752053Assignment : Hypothesis Testing -23.1759908644 13.398570599Dataset 2 20 0.1Part (a) 207.20202.13196.93198.16197.74198.15 0.8207.65203.68197.13197.06196.60197.55 208.93202.22198.63197.09197.40198.36 207.51201.32197.97198.31198.04198.78 206.02200.07196.67199.85199.05200.31 205.84199.09197.67198.40200.32199.29 204.36198.89196.90197.34199.11200.46 Part (b) 203.64197.56198.07198.70198.13202.00 203.23198.43199.61197.65198.25200.55 198.56199.07199.70199.13203.00204.23 6.9546138291-14.2408215775-36.0176781469-30.8666184359-32.6388550472-30.9335754424 8.8455546286-7.7852072629-35.2120054224-35.4989160217-37.4147752053-33.4140912344 14.1984112513-13.8962725431-28.9263544042-35.3555572248-34.0663805275-30.0421974204 8.2637367734-17.6578154853-31.6681723018-30.2389660677-31.387474862-28.274006656 2.0212369411-22.8856482931-37.1113268042-23.7880614796-27.1521956225-21.880094171 1.2742778891-26.9762038377-32.9126083269-29.8730791501-21.8507030202-26.1353097278 -4.9179951627-27.8245078799-36.1352421761-34.3286613999-26.8844859805-21.2517703254 -7.9549632153-33.4057411699-31.253486167-28.624322408-31.0099289575-14.7859664342 -9.6324886749-29.7519662508-24.7888156185-33.0066980522-30.5130282256-20.8902981682 -14.2408215775-36.0176781469-30.8666184359-32.6388550472-30.9335754424-24.8390685042 &"Helvetica Neue,Regular"&12&K000000&P Dataset 3 -37.4147752053Assignment : Hypothesis Testing -23.1759908644 13.398570599Dataset 3 20 0.112 1500List (a) 101571.961521.341469.331481.631477.401481.471500.52 0.81576.481536.761471.251470.571465.991475.551470.84 1589.261522.161486.271470.911473.991483.601504.37 1575.091513.181479.721483.131480.391487.82 List (b) 1548.181488.691454.721486.541478.501491.101477.71 1546.391478.921464.751472.011491.171480.931489.03 1531.611476.901457.051461.361479.141492.601472.54 1524.351463.571468.711474.991469.291508.041484.82 1520.351472.291484.151464.521470.481493.46 6.9546138291-14.2408215775-36.0176781469-30.8666184359-32.6388550472-30.9335754424 8.8455546286-7.7852072629-35.2120054224-35.4989160217-37.4147752053-33.4140912344 14.1984112513-13.8962725431-28.9263544042-35.3555572248-34.0663805275-30.0421974204 8.2637367734-17.6578154853-31.6681723018-30.2389660677-31.387474862-28.274006656 2.0212369411-22.8856482931-37.1113268042-23.7880614796-27.1521956225-21.880094171 1.2742778891-26.9762038377-32.9126083269-29.8730791501-21.8507030202-26.1353097278 -4.9179951627-27.8245078799-36.1352421761-34.3286613999-26.8844859805-21.2517703254 -7.9549632153-33.4057411699-31.253486167-28.624322408-31.0099289575-14.7859664342 -9.6324886749-29.7519662508-24.7888156185-33.0066980522-30.5130282256-20.8902981682 &"Helvetica Neue,Regular"&12&K000000&P Dataset 4 -37.1113268042 14.1984112513Assignment : Hypothesis Testing 20Dataset 4 0.1 0.8Resistance: 14 0.8Motor runningMotor not running 0.8615.7215.12-26.97620383770.2 0.915.8015.18-27.82450787990.18 6.9546138291116.0015.27-33.40574116990.07 8.84555462860.8815.7615.10-29.75196625080.14 14.19841125130.7615.5214.74-36.01767814690.02 8.26373677340.7515.5014.74-35.21200542240.04 2.02123694110.6315.2614.62-28.92635440420.16 1.27427788910.5715.1414.45-31.66817230180.11 -4.91799516270.5415.0814.28-37.11132680420 -7.95496321530.4514.9014.18-32.91260832690.08 -9.63248867490.5715.1414.36-36.13524217610.02 -14.24082157750.4514.9014.21-31.2534861670.11 -7.78520726290.3814.7614.20-24.78881561850.24 -13.89627254310.2814.5613.88-30.86661843590.12 -17.65781548530.214.4013.63-35.49891602170.03 -22.88564829310.1814.3613.59-35.35555722480.03 -26.97620383770.0714.1413.47-30.23896606770.13 -27.82450787990.1414.2813.74-23.78806147960.26 -33.40574116990.7215.4414.78-29.87307915010.14 -29.75196625080.7215.4414.69-34.32866139990.05 &"Helvetica Neue,Regular"&12&K000000&P Dataset 5 -53.3841605322 -3.7575430833Assignment : Hypothesis Testing 0.8 28Dataset 5132 0.20.15 4.8 62AdditiveYield -8.0764593608177119.68-29.70893359550.48 0.20.824637077974.37119.26-38.12655717960.31 62.20.873099097975.1119.78-31.56298442770.44 -3.75754308330.938182966876.07119.29-34.79876530310.37 -12.46021173240.782353857873.74118.88-43.09633574840.21 -10.05520560480.721758012872.83119.49-37.82497821430.31 -6.82531334140.65368474171.81118.92-45.56678519640.16 -14.55858492040.595816637470.94118.72-48.99796024830.09 -17.56575173870.435381757568.53119.08-48.9121832170.09 -20.94399795980.262518020365.94119.04-53.38416053220 -23.81579619870.407566485768.11119.29-47.25492088740.12 -31.7776366070.56202108470.43119.85-38.14650388360.31 -40.35627916650.573998124770.61120.5-30.8528053020.45 -33.15801446340.471168617869.07120.06-37.76985777580.31 -25.4929551994 -24.8985751799 -30.0016557832 &"Helvetica Neue,Regular"&12&K000000&P Dataset 6 Assignment : Hypothesis Testing 15Dataset 6 0.12 G1G2G3 A101314 B18116 SchoolC162017 D122513 E52214 -2.2880010846 9.3864597430.82.4711910153.2754334611 11.67446082765.46096037181.7316334316-1.3183792526 4.42148107486.59452430933.1423297427 2.2803120249.386459743-0.460755274 -2.28800108466.3819096050.8246918139 3.2754334611 5.2031240217 0.7540790008 &"Helvetica Neue,Regular"&12&K000000&P Reference 3 678 Student numbersLast3Seed value B0014420420410.8 B001442242241 B001454764761 B001489359350 B001464794791 B001406626621 B001444634631 B001468378370 B001463093091 B001432192191 B00144085851 B001423533531 B001444144141 B001398008000 B001473473471 B001433073071 B00133044441 B001451101101 B001019679670 B001414644641 B000515705701 B001416506501 B001488828820 B001563043041 B001362762761 B001484884881 B001205855851 B001452952951 B001423003001 B001493463461 B001465345341 B001444484481 B000792332331 B00132032321 B001424584581 B00149056561 B001435825821 B001454394391 B001471961961 B001466276271 B001464554551 B001418148140 B001363223221 B001419019010 B001407247240 B001483283281 B001458538530 B0014600771 B001464634631 B001235225221 B001418788780 B001419839830 B001425035031 B001453673671 B001399759750 B00146051511 B001481461461 B001367657650 B001342102101 B001489599590 B001488348340 B001485725721 B00141067671 B001386406401 B001448638630 B001458768760 B00134039391 B001429569560 B00137073731 B001349699690 B00141010101 B001466886880 B001361871871 B001388828820 B001491121121 B001372822821 B001466546541 B000983733731 B001461761761 B00140024241 B001409729720 B001473393391 B001453123121 B00147010101 B00143095951 B001426106101 B001421981981 B001404304301 B001437547540 B001358418410 B001353113111 B001329469460 B001028528520 B001416766761 B001377707700 B00142018181 B001442552551 B001455025021 B000718388380 B001411111111 B00147030301 &"Helvetica Neue,Regular"&12&K000000&P
Answered 2 days AfterMay 03, 2023

Answer To:

Baljit answered on May 06 2023
25 Votes
Question 1:-
    Groups
    Frequencies
    Grouped Mean
    Cumulative Frequency
    300
    to
    307
    12
    303.5
    12+0=12
    307
    to
    314
    18
    
    12+18=30
    314
    to
    321
    44
    
    30+44=74
    321
    to
    328
    88
    =324.5
    74+88=162
    328
    to
    335
    86
    =331.5
    162+86=248
    335
    to
    342
    41
    
    
248+41=289
    342
    to
    349
    15
    
    289+15=304
    349
    to
    356
    9
    =352.5
    304+9=313
1.

2.
Mean:
Standard Deviation:
s
Sd=
Median:-
Middle Value (N/2)=313/2=156.5
Now Middle Values lies between 321-328
Here l is lower limit of median class =321
Cf is cumulative frequency of class preceding the median class
F is frequency of median class =88
h is class size =7
So
327.5625
First Quartile:
N/4 i.e 313/4=78.25 lies between interval 321-328
*7=321.33
Third Quartile:
3N/4=4*313/4=234.75 lies in interval 328-335
*7=333.92
3.
As we can see from the mean and median which are equal to each other So that mean data distribution is symmetric. Same we can see it from the frequency polygon and histogram that data is symmetric.
    
4. The mean and standard deviation for grouped data are calculated by treating each interval as a single value (the midpoint) and using the same formulas as for ungrouped data. The formula for mean is the weighted average of the midpoints, where the weights are the frequencies. The formula for standard deviation is the square root of the weighted average of the squared deviations from the mean, where the weights are the frequencies.
5. If a continuous data is follow normal distribution then 68.2%, 95.4%, and 99.7% observations are lie between
mean ± 1 SD, mean ± 2 SD, and mean ± 3 SD, respectively.
i. Now 68.2% observation is =0.682*313=213.466
Which is lies in interval 328-335
Now
Mean ± 1 SD=327.4744 ± 10.46 is 317.0144 to 337.93
So 68.2 % observation lies between mean ± 1 SD
ii. Now 95.4% observation is =0.954*313=298.602
Which is lies in interval 342-349
Now
Mean ± 2 SD=327.4744 ± 2* 10.46 is 306.5544 to 348.4944
So 95.4 % observation lies between mean ± 2 SD
iii. Now 99.7% observation is =0.997*313=312.06
Which is lies in interval 349-356
Now
Mean ± 3 SD=327.4744 ± 3* 10.46 is 296.09 to 358.8544
So 99.7 % observation lies between mean ± 3 SD
Hence our distribution is normal distribution.
Question 2:-
Part(a)
1.
Sample Mean (
Sample Standard Deviation,
Sample Standard Deviation,
Sample Standard Deviation,
We are using term sample mean and standard deviation because it is the mean and standard deviation of a small sample of the Population.
2. We will use Z-test Hypothesis
To calculate Z
Here
: Sample Mean
Standard deviation
So for calculation of the Z value we need sample mean
3. Now
For significance level
Z Critical Value for ,
Now Since
Hence we failed to Reject Null Hypothesis So we don’t have sufficient evidence to claim that mean is not 200mm
Part(b)
1. Sample Mean (
Sample Standard Deviation,
Sample Standard Deviation,
Sample...
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