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# Math IA

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This is my internal assessment for my IB Math Studies class where I did an investigation on the correlation between height and weight of students and whether the two are dependent of each other. This …

This is my internal assessment for my IB Math Studies class where I did an investigation on the correlation between height and weight of students and whether the two are dependent of each other. This investigation is based on data I collected from students within my class.

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• 1. 1 IB Math Studies Internal Assessment: Height and Weight of Boys and Girls Tanessa Sirimongkarakorn School name: International School of Bangkok Date: November 2010 Course: IB Math Studies
• 2. 2 INTRODUCTION There are many shapes and sizes of boys and girls, in which their height and weight vary among the population. Because of the variety of size, I thought it would be interesting to look at whether greater height results in greater weight. This investigation examines the height versus the weight of boys and girls in order to see if height has a significant effect on weight. For this investigation, I will survey the height and weight of 25 boys and 25 girls. I will then work to find the correlation between heights and weight both of the boys and the girls. To find the correlation as well as the correlation strength of the data, I will use Pearson’s correlation coefficient formula. A Chi Squared test will also be applied in order to find whether or not the height variable is dependent on the weight variable. I chose to do the investigation on this topic because of my interest in the human body. With many concerns and pressure among today’s society regarding body types and sizes, I wanted to find if there is a reason for the varying sizes as well as if gender is a factor responsible for the differences of height and weight. HYPOTHESIS There will be a positive correlation between height and weight; the taller the person is, the greater their weight will be. From my observations, I have noticed that typically people of greater height have greater weights. DATA COLLECTION In order to collect my data, I surveyed 25 boys and 25 girls from my school of my grade. I asked the groups for their height and their weight and recorded the data accordingly by gender. I recorded their heights in inches and their weights in kilograms. I then separated the data of heights into respective groups by inches, and found the average weight among those groups. This is showed on the table below.
• 3. 3 Table 1: Data for boys’ heights and average weights Height (inches) Number of Boys of Height Average Weight 65 3 67.3 66 2 70 68 4 66 69 5 71.8 70 4 73.75 72 3 82.3 73 1 86 74 1 62 75 1 85 83 1 70 Table 2: Data for girls’ heights and average weights Height (inches) Number of Girls of Height Average Weight 57 1 65 60 1 44 61 1 50 62 5 51.8 63 2 53 64 2 56 65 5 52.6 66 4 52.2 67 2 62 68 2 57.5 Table 1 and Table 2 both show the data where the height is categorized by inches while also showing the average weight of each group of height. However, a problem that is present within the tables is the inconsistency of people per height group, which leads to an inconsistent weight average. For example, in table 2, there are 5 people with a height of 62 inches and with a height of 65 inches but within the height groups of 57, 60, and 61 inches; there is only one person each of those heights. Because of these inconsistencies, the scatter plot may turn out to be an inaccurate visual.
• 4. 4 DATA ANALYSIS The tables below show the data of both boys and girls with the average weight rounded to the nearest kilogram. Table 3: Boys’ Data Height (x) Average Weight Rounded to the Nearest Kilogram (y) 65 67 66 70 68 66 69 72 70 74 72 82 73 86 74 62 75 85 83 70 Table 4: Girls’ Data Height (x) Average Weight Rounded to the Nearest Kilogram (y) 57 65 60 44 61 50 62 52 63 53 64 56 65 53 66 52 67 62 68 58
• 5. 5 Diagram 1: Diagram 2: These scatter plots show the height group to the average weight of that height group rounded to the nearest kilogram as well as with outliers removed. The boys’ data is represented on the top diagram, while the girls’ is represented on the bottom diagram. Outlier points which have been removed from the boys’ data are points (74, 62) and (83, 70). Point (57, 65) from the girls’ data was also considered an outlier and was removed. 0 10 20 30 40 50 60 70 80 90 100 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 Boys' Height to Average Weight Average Weight for Boys (kilograms) 0 10 20 30 40 50 60 70 80 90 100 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 Average Weight for Girl s (kilograms) Average Weight for Girls (kilograms)
• 6. 6 The relationship of the boy’s data seems to be a positive linear one and also shows moderate correlation strength. The girls’ data also seems to be a positive linear relationship. However, it shows stronger correlation strength than the boys’. Pearson’s correlation coefficient formula will be used in order to discover the real strength of correlation. THE CORRELATION COEFFICIENT Below is the equation for the r-value, which will then be used to determine the strength according to Pearson’s chart. yyxx nn yxnxy r 2222 )(      Values of the formula: Boys Girls Mean of x 69.8 64.04 Sum of x 1745 1601 Sum of x2 122159 102695 Mean of y 72.4 54 Sum of y 1810 1350 Sum of y2 132052 73348 Sum of xy 126580 86510 Using these values, I plugged them into the r-value equation. Below shows the set up of the equation. Boys: Boys’ r-value= .4028502028 Girls:
• 7. 7 Girls’ r-value= .2047589067 The chart below signifies the correlation strength of relationships according to their r value. According to the chart, the boys’ data of height and weight have a positive weak correlation. Additionally, the girls’ data shows a very weak positive correlation. Correlation Strength Chart value Strength 0< <0.25 Very weak correlation 0.25< <0.50 Weak correlation 0.50< <0.75 Moderate correlation 0.75< <0.90 Strong correlation 0.90< <1.0 Very strong correlation =1 Perfect correlation FINDING CHI SQUARED Next, the Chi Squared Test needs to be done in order to test the independence of our data. This will show whether a person’s weight is depended on their height. This test will also reveal the results regarding the null hypothesis, in which will either be accepted or rejected. To perform the Chi Squared test, tables of observed values and expected values will be needed. To begin, the table of expected values must be found. These values are found by multiplying the total sum of the corresponding column and the total sum of the corresponding row, and dividing the product by the bottom right value of the table. These values are rounded to three significant figures.
• 8. 8 For example, the first value of the table (.391) was found by calculating (3 x 3) / 23. Boys: Height (inches) 65 66 68 69 70 72 73 75 Total 67 .391 .261 .522 .652 .522 .391 .130 .130 3 70 .261 .174 .348 .435 .348 .261 .087 .087 2 66 .522 .348 .696 .870 .696 .522 .174 .174 4 72 .652 .435 .870 1.09 .870 .652 .217 .217 5 74 .522 .348 .696 .870 .696 .522 .174 .174 4 82 .391 .261 .522 .652 .522 .391 .130 .130 3 86 .130 .087 .174 .217 .174 .130 .043 .043 1 85 .130 .087 .174 .217 .174 .130 .043 .043 1 Total 3 2 4 5 4 3 1 1 23 Girls: Height (inches) 60 61 62 63 64 65 66 67 68 Total 44 .042 .042 .208 .083 .083 .208 .167 .083 .083 1 50 .042 .042 .208 .083 .083 .208 .167 .083 .083 1 52 .208 .208 1.04 .417 .417 1.04 .833 .417 .417 5 53 .083 .083 .417 .167 .167 .417 .333 .167 .167 2 56 .083 .083 .417 .167 .167 .417 .333 .167 .167 2 53 .208 .208 1.04 .417 .417 1.04 .833 .417 .417 5 52 .167 .167 .833 .333 .333 .833 .667 .333 .333 4 62 .083 .083 .417 .167 .167 .417 .333 .167 .167 2 58 .083 .083 .417 .167 .167 .417 .333 .167 .167 2 Total 1 1 5 2 2 5 4 2 2 24
• 9. 9 Now we apply our data to the Chi Squared formula, which is: The O value represents the observed value, while the E value represents the expected value. The results for the boys came to be 4.6050386, and the girls’ chi square value came to be 3.664213403. Rounded to three significant figures, the boys’ chi square value is 4.61 and the girls’ value is 3.66. Lastly, the formula of degrees of freedom needs to be applied as well in order to complete our test of independence. The formula for degrees of freedom is: Df= (number of rows-1)(number of columns-1) The degree of freedom for the boys as well as the girls is 9. To find out the results of dependence of the two variables, following diagram is of the Chi Square Distribution, which is a table to be referred to in order to determine if our hypothesis is accepted or rejected. V represents the degree of freedom.
• 10. 10 If the Chi Squared value is greater than the critical value for a 5% significant level, the null hypothesis is rejected, proving the dependence of the two variables. If the Chi Squared value is less than the critical value for a 5% significant level, the null hypothesis is then accepted; meaning there two variables are independent of each other. CONCLUSION Through this process we were able to find the correlation strength as well as the strength of the dependence. Pearson’s correlation coefficient formula led us to find that the correlation strength of the boys’ height and weight had positive yet weak correlation strength and that the girls’ data also showed to be a positive correlation yet a very weak one. According to our results from the Chi Square test, we found the chi square value of the boys to be less than the k value from the Chi Square Distribution table (4.61<16.919). Similarly, the girls’ Chi Square value also was found to be less than the k value (3.66<16.919). The null hypothesis is that there is no correlation between the two variables, and the results of these tests have confirmed that the null hypothesis is accepted; that there is no association between the variables of height and weight.
• 11. 11 POSSIBLE ERRORS There are many factors throughout this investigation that could have led to invalid reasoning. One is that through the data collecting process, it is possible that the information regarding people’s heights and weights were not completely accurate. This is because during the surveys, the participants were simply asked for their information, and there were no further actions towards confirming if the data gathered is accurate. This accounts for many of the values, which are not considered to be controlled, and in result leads to inaccurate tables, graphs, and inferences. Also, this investigation does not account for the rest of the population; only 50 people were surveyed and studied which evidently cannot hold credibility for the whole population. WORKS CITED Coad, Mal. Mathematics for the International Student: Mathematical Studies SL for Use with IB Diploma Programme. Adelaide Airport, SA: Haese & Harris, 2004. Print.