Quantitative Data Analysis Assignment for Business Research

BRDA Assignment - 2 Business Research and Data Analysis (Conestoga College) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university BRDA Assignment - 2 Business Research and Data Analysis (Conestoga College) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by Lekshmy G (lekshmygangadharan@gmail.com) lOMoARcPSD|37560762

BUS8375 Business Research and Data Analysis Assignment 2- Quantitative Data Analysis Name: Dhruvish Patel Conestoga DTK Campus Dr Moses Moussa 22/4/2023 Downloaded by Lekshmy G (lekshmygangadharan@gmail.com) lOMoARcPSD|37560762

Assignment 2 – Quantitative Data Analysis Review the data collected from 24 post-secondary students, including age, exam mark, essay mark, gender, year in college and IQ. Using this data, complete the following: 1) Create a FREQUENCY DISTRIBUTION table and Draw a HISTOGRAM to present the IQs (use 5 ranges) in five intervals. 3 MARKS – SHOW YOUR CALCULATIONS IN DETAIL IQ range Frequency Mid Point RELATIVE FREQUENCY Cumulative Frequency 65-87 8 76 0.34 8 87-109 7 98 0.29 15 109-131 7 120 0.29 22 131-153 1 142 0.04 23 153-175 1 164 0.04 24 Total 24 1 Downloaded by Lekshmy G (lekshmygangadharan@gmail.com) lOMoARcPSD|37560762

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2) Draw a PIE CHART including percentages and degrees summarizing the Gender . 3 MARKS – SHOW YOUR CALCULATIONS IN DETAIL 3) Draw a scatterplot showing Essay Marks on the X-axis and Age on the Y-axis. 1 MARK – SHOW YOUR CALCULATIONS IN DETAIL 18 20 22 24 26 28 30 32 34 36 0 10 20 30 40 50 60 70 80 90 100 Age Essay Marks (%) In summary, there appears to be a positive correlation between Age and Essay marks, suggesting that the higher the Age the higher the Essar mark. Gender Frequency Proportion X/Y Male 11 0.46 Female 10 0.42 Other 3 0.12 Total 24 1 Downloaded by Lekshmy G (lekshmygangadharan@gmail.com) lOMoARcPSD|37560762

4) Calculate the Mean and find the Median and Mode for the Years in College? 1 MARK – SHOW YOUR CALCULATIONS IN DETAIL Years In College = 2,1,4,1,3,3,2,4,4,3,1,2,1,1,3,2,3,4,3,3,2,3,2,3 Mean = Sum of all years in college divided by the total frequency 2+1+4+1+3+3+2+4+4+3+1+2+1+1+3+2+3+4+3+3+2+3+2+3 = AVERAGE (B4:B27) case of the Excel attached 60/24 = 2.5 Median = Arrange the dataset from the lowest to the largest of vice versa and find the middle number 1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,4,4,4,4 =MEDIAN (I4:I27) case of the Excel attached (3+3) / 2 3 Mode = This is the number that appears most times in the data set 1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,4,4,4,4 From the above data set, 3 appears the most times (9). =MODE (R4:R27) case of the Excel attached. Downloaded by Lekshmy G (lekshmygangadharan@gmail.com) lOMoARcPSD|37560762

5) Calculate the STANDARD ERROR and SKEWENESS (identify what type) for Exam Marks (Consider n = 24). 7 MARKs – SHOW YOUR CALCULATIONS IN DETAIL Step 1: find the mean for Exam Marks 86+84+86+81+80+65+75+90+78+89+72+81+75+76+80+75+85+79+91+89+90+87+95+ 76 SUM = 1965/24 81.875 STEP 2: Finding the marks error/deviation from the mean. Subtract the mean from each IQ score so as to get deviations from the mean Exam mark Deviation from mean Square each deviation from the mean 86 86-81.875=4.125 17.015625 84 84-81.875=2.125 4.515625 86 86-81.875=4.125 17.015625 81 81-81.875=-0.875 0.765625 Downloaded by Lekshmy G (lekshmygangadharan@gmail.com) lOMoARcPSD|37560762

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80 80-81.875=-1.875 3.515625 65 65-81.875=-16.875 284.715 75 75-81.875=-6.875 47.265625 90 90-81.875=8.125 66.015625 78 78-81.875=-3.875 15.015625 89 89-81.875=7.125 50.765625 72 72-81.875=-9.875 97.515625 81 81-81.875=-0.875 0.765625 75 75-81.875=-6.875 47.265625 76 76-81.875=-5.875 34.515625 80 80-81.875=-1.875 3.515625 75 75-81.875=-6.875 47.265625 85 85-81.875=3.125 15.015625 79 79-81.875=-2.875 4.515625 91 91-81.875=9.125 97.515625 89 89-81.875=7.125 50.765625 90 90-81.875=8.125 66.015625 87 87-81.875=5.125 34.515625 95 95-81.875=13.125 172.265625 76 76-81.875=-5.875 34.515625 Downloaded by Lekshmy G (lekshmygangadharan@gmail.com) lOMoARcPSD|37560762

1965 1212.574375 Step 4: Finding the sum of the squares. 1212.574 Step 5. Finding the variance Variance = the sum of all squares divided by n-1, where n is the sample size (24) 1212.574/ (24-1) 1212.574/23 Variance = 52.720608 Step 6: then find the square root of the variance. Standard Deviation = √ 52.720608 SD= 7.11 thus, therefore, conclude that each exam mark deviates from the mean by 7.11 thus showing a positive skewness. Step 6. Calculate Skewness Skewness = ∑ N i (X i – X ) 3 / (N-1) * σ 3 Skewness = (3 * (mean-median))/standard deviation Mean= 81.88 Downloaded by Lekshmy G (lekshmygangadharan@gmail.com) lOMoARcPSD|37560762

Median= 81.00 Standard deviation= 7.11 81.88 - 81.00 =0.88 0.88 * 3 = 2.64 2.64/ 7.11 = -0.26979 A positive skewness . It indicates that the distribution of marks has a longer tail on the right side (positive side) of the distribution. Standard Deviation, σ=∑ ni=1 (x i −x ¯¯¯ ) 2 n−−−−−−−−−−− −√ �=∑�=1�(��−�¯)2� Standard Deviation, σ: 7.0374740970512 Count, N: 24 Sum, Σx: 1965 Mean, μ: 81.875 Variance, σ 2 : 49.526041666667 Steps σ 2 = Σ(x i - μ) 2 N = (86 - 81.875) 2 + ... + (76 - 81.875) 2 Downloaded by Lekshmy G (lekshmygangadharan@gmail.com) lOMoARcPSD|37560762

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24 = 1188.625 24 = 49.526041666667 σ = √49.526041666667 = 7.0374740970512 s=∑ ni=1 (x i −x ¯¯¯ ) 2 n−1−−−−−−−−−−−−√ � =∑ � =1 � ( �� − � ¯)2 � −1 s=SSn−1−−−−−√ � = �� −1 s=1188.62524−1−−−−−−−−√ � =1188.62524−1 s=1188.62523−−−−−−−−√ � =1188.62523 s=51.679348−−−−−−−−√ � =51.679348 s=7.1888349 Variance = σ 2 = Σ ( x i − μ ) 2/N The population standard deviation, the standard definition of σ , is used when an entire population can be measured, and is the square root of the variance of a given data set. Take the square root of the population variance to get the standard deviation. Population standard deviation = √ σ 2 Downloaded by Lekshmy G (lekshmygangadharan@gmail.com) lOMoARcPSD|37560762

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brda-assignment-4 quantitative analysis

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