a1s - MATH 1131 3.0 - Fall 2011 Solution for Assignment 1...

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MATH 1131 3.0 - Fall 2011 Solution for Assignment 1 Question 1: a) The 5-number summary from MINITAB are: Male: 410.00 490.75 504.50 527.00 554.00 Female: 401.00 480.50 494.00 514.00 543.00 Using R, we have Male: 410.0 493.8 504.5 526.2 554.0 Female: 401.0 482.8 494.0 513.0 543.0 Using hand calculation, we have Male: Q1 is between 484 and 493, Median is between 504 and 505, and Q3 is between 527 and 527 Therefore the 5-number summary is: 410.0 488.5 504.5 527.0 554.0 Female: Q1 is between 476 and 482, Median is between 494 and 494, and Q3 is between 513 and 517 Therefore the 5-number summary is: 410.0 479.0 494.0 515.0 543.0 Any of the above answer is acceptable. b) Many graphical presentations are and the following is the back-to-back stemplot (Leaf Unit = 1): Female Stem Male 1 40 41 0 1 42 7 43 44 973 45 46 2 60 47 049 77652 48 4 64421 49 36899 8840 50 03457 7733 51 3478 034 52 477 53 3467 54 55 24 The distributions of males’s scores and females’ scores have more or less the same shape (skewed with a long left tail). The six highest scores are from males and the lowest score is from female. However, the difference in scores does not seem like it is a big difference. I will say that they have very similar performances.
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OR We can use the side-by-side boxplots. Females Males 560 540 520 500 480 460 440 420 400 Data Boxplot of Males, Females From the plot, the performances are more or less the same (even though females’ scores seem to be slightly lower than males’ score, but they are not significantly lower because the boxes overlapped). c) Since the distribution of males’ scores does not look like it is bell-shaped, we will be using the inner fences and outer fences to determine if outliers exist or not. I am going to use the MINITAB calculations shown in part (a) to calculate the
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a1s - MATH 1131 3.0 - Fall 2011 Solution for Assignment 1...

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