Estimation and testing of Hypotheses

Organisation of raw data* Back to back stem and leaf diagram* Cumulative frequency tables* Cumulative frequency graphs (graph paper)* Box-plot diagram (graph paper)* Standard deviationAveragesMale estimates/ female estimates* Median* Lower quartile* Upper quartile* Inter quartile range* Mean* Mode* RangeConclusionHypothesis two: -Raw data,Organisation of raw data,* Back to back stem and leaf diagram* Histogram/frequency polygon tables* Histogram graphs/frequency polygons (graph paper)AveragesUnder 18 estimates/ over 18 estimates* Median* Mean* Mode* RangeConclusionEvaluationPlan:Aim: to investigate and prove (or not be supported by the data obtained) two hypotheses.Problem: Sarah asked students to estimate the length of a line and the size of an angle.Using two hypotheses, design and carry out an experiment to test these.Hypotheses: 1) Girls are better at estimating lengths than boys.2) Boys over the age of 18 are better at estimating the size of an anglethan boys under 18.First hypothesisSample population and methodWhen conducting the experiment I will attempt to ask every third person to estimate the sizes of the line and the angle. I will collect thirty results for each gender.The rule of selection will be every third person (of either gender) will be asked to estimate the length of the line and the size of the angle.Using this sample technique I expect the results collected to be random.The people asked to participate in the investigation will be students at the college; this will skew the age range of the participants. I will be collecting the data at college, as this is convenient and will lessen the impact on my other study activities.If I were to collect data at other colleges or out in the general public the results may differ somewhat, but at this initial stage, I will declare that examination of this as a factor is out of scope of this exercise..First hypothesisRaw data:MaleData itemLength (cm)014.5025.5035.5045054.5065.5074.5084.9094.5104.5113.9124133.5143.4154.5164.5174.9183.9193.9204.5214.9224.5234.9245254.5264.5274.5285395.5304.5FemaleData itemLength (cm)314.5324.5334.5345354.5365373.9384.5394.8405414.5424.5434.9444.3454.5465474.5484494.5504.5514525.5535544.5554.5563.9574583.9593.9604Organisation of data for 1st hypothesisBack to back stem and leaf diagram for estimated lengthsBoys Girls99954 3 9999999955555555555550 4 00003555555555555895555000 5 0000005Cumulative frequency tables for length estimatesMaleClassTallyFrequencyCumulative frequency3.0? x < 3.51113.5? x <4.01111454.0? x <4.51164.5? x <5.01111 1111 1111 1117235.0? x <5.51113265.5? x <6.01111430FemaleClassTallyFrequencyCumulative frequency3.5? x <4.01111444.0? x <4.51111594.5? x <5.01111 1111 111115245.0? x <5.511115295.5? x <6.01130MalesMedian1/2 x (30+1)= 15.5Median length = 4.76 cmLower Quartile1/4 x (30+1)= 7.75LQ length = 4.54 cmUpper quartile3/4 x (30+1)= 23.25UQ length = 5cm?Interquartile range = 5 – 4.54= 0.46 cmMean (from back to back stem and leaf diagram)Mean = ?fx?f= 137.730= 4.59 cmMode4.5 cmRange5.5-3.4= 2.1 cmFemalesMedian1/2 x (30+1)= 15.5Median length = 4.71 cmLower quartile1/4 x (30+1)= 7.75LQ length = 4.41 cmUpper quartile3/4 x (30+1)= 23.25UQ length = 4.91 cm? Interquartile range = 4.91 – 4.41= 0.08 cmMean (from back to back stem and leaf diagram)Mean = ?fx?f=134.630=4.49 cm (2 d.p)Mode=4.5 cmRange=5.5-3.9= 1.6 cmStandard deviationThis is representative of the “Spread” of estimates about the mean. It is calculated by subtracting the mean from each data value, squaring these values, and adding these squared values together. This sum of the squared values is divided by one less than the number of original data values, and the square root of this division is the Standard Deviation.BOYSmean =4.59estimateestimate – mean(estimate – mean)^24.5-0.090.00815.50.910.82815.50.910.828150.410.16814.5-0.090.00815.50.910.82814.5-0.090.00814.90.310.09614.5-0.090.00814.5-0.090.00813.9-0.690.47614-0.590.34813.5-1.091.18813.4-1.191.41614.5-0.090.00814.5-0.090.00814.90.310.09613.9-0.690.47613.9-0.690.47614.5-0.090.00814.90.310.09614.5-0.090.00814.90.310.096150.410.16814.5-0.090.00814.5-0.090.00814.5-0.090.008150.410.16815.50.910.82814.5-0.090.0081Sum of (estimate – mean)^28.687Divide by 290.299551724Root0.547313187Standard Deviation :0.547313187Girlsmean =4.486666667estimateestimate – mean(estimate – mean)^24.50.0133333330.0001777784.50.0133333330.0001777784.50.0133333330.00017777850.5133333330.2635111114.50.0133333330.00017777850.5133333330.2635111113.9-0.5866666670.3441777784.50.0133333330.0001777784.80.3133333330.09817777850.5133333330.2635111114.50.0133333330.0001777784.50.0133333330.0001777784.90.4133333330.1708444444.3-0.1866666670.0348444444.50.0133333330.00017777850.5133333330.2635111114.50.0133333330.0001777784-0.4866666670.2368444444.50.0133333330.0001777784.50.0133333330.0001777784-0.4866666670.2368444445.51.0133333331.02684444450.5133333330.2635111114.50.0133333330.0001777784.50.0133333330.0001777783.9-0.5866666670.3441777784-0.4866666670.2368444443.9-0.5866666670.3441777783.9-0.5866666670.3441777784-0.4866666670.236844444Sum of (estimate – mean)^24.974666667Divide by 290.17154023Root0.414174154Standard Deviation :0.414174154ConclusionWhen comparing the mean and the standard deviation of the estimates for the females and males I find that although the males mean is closer to the actual length of the line, the girls have a smaller standard deviation. This indicates that although further from the actual length of the line they are more consistent as a group.The range for the females also supports this, it is a 0.5 cm difference between the two ranges.Conceivably if the sample population had been given more specific guidelines for estimation of the line length then the results may have been more consistent and closer to the actual length. Examples of this may be to ask the sample population to estimate the length of the line to two decimal places or to specify that the line begins and ends before the ink from the other two lines closes the length.Ultimately, I feel that I set out on this investigation without limiting enough of the degrees of freedom necessary to be able to point to the resultant evidence as either supporting or differing with the hypothesis.Hypothesis twoMales over the age of 18 are better at estimating the size of an angle than boys under 18.Raw dataData itemAgeAngle ( ? )011845021530031837042040051742061740071750081545091445101845111745121939131735141945151740161840171745181560191455201845212140222139231845241940251640261845271545281545291840302043311643321535331630341633351444361445371644381845392042402139412029421645431947441629451845462045472147482030491747501654511641521645531724542030552144562245572042581739591937601850Organisation of dataBack to back stem and leaf diagramUnder 18 18 or over94 2 9955300 3 0077999755555555544321000 4 000002234555555555577540 5 00 6Histogram/ frequency polygonDataUnder 18classFrequency20? x <30230? x <40640? x <501850? x <60360? x <70118 and overclassFrequency30? x <40140? x <50750? x <602160? x <701AveragesUnder 18Mean= 126030= 42?Median= 42?Mode= 45?Range= 60-24=36 ?Over 18Mean=124130= 41.37? (2 d.p)Median= 42.5?Mode= 45?Range=50-29= 21?Boys<18mean =40.42029-11.4129.962030-10.4108.162030-10.4108.161837-3.411.561937-3.411.561939-1.41.962039-1.41.962139-1.41.962139-1.41.961840-0.40.161840-0.40.161940-0.40.162040-0.40.162140-0.40.1620421.62.5620421.62.5621443.612.9618454.621.1618454.621.1618454.621.1618454.621.1618454.621.1618454.621.1618454.621.1619454.621.1620454.621.1622454.621.16Standard Deviation :21.1619476.643.5621476.643.5618509.692.16Ageestimateestimate – mean(estimate – mean)^2Sum of (estimate – mean)^2548.88Divide by 2918.92689655Root4.350505321Boys>18mean =41.5Ageestimateestimate – mean(estimate – mean)^22029-12.5156.252030-11.5132.252030-11.5132.251837-4.520.251937-4.520.251939-2.56.252139-2.56.252139-2.56.251840-1.52.251840-1.52.251940-1.52.252040-1.52.252140-1.52.2520420.50.2520420.50.2520431.52.2521442.56.2518453.512.2518453.512.2518453.512.2518453.512.2518453.512.2518453.512.2518453.512.2519453.512.2520453.512.2522453.512.2519475.530.2521475.530.2518508.572.25Sum of (estimate – mean)^2755.5Divide by 2926.05172414Root5.104088963Standard Deviation :5.104088963ConclusionUnfortunately the data collected is only marginally diverse. But from this there are data items that do show a fragment of variation between them.The actual angle was 37.5?, although minute, the over 18’s were closer to the actual angle with a mean of 41.37? (2d.p). However both have different ranges with a difference of 15?.The histograms showed the distribution of the frequencies. They do show a slight difference on the position of the modal class.With the data and data presentation that I have I cannot in good confidence decide weather it invalidates or proves my hypothesis.EvaluationIf I had the chance to repeat the investigation I would have liked to improve my hypotheses by making them more precise and accurate to allow maximum accuracy towards proving a hypothesis. This would have endorsed me to come to a solid conclusion.As mentioned in the beginning of the investigation I only conducted the experiment inside the college, this did limit the age range of the sample population and may have contributed to the loss of a valid and inclusive conclusion.

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