M & M Project Think! Crunch those numbers! Answer!
Chapters 1-2 Exploring Data and Describing Location in a Distribution
Univariate Data: Length Stemplot and Frequency Table Stem (Units Digit) 0 1 1 Leaf (Tenths digit) 6 8 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 5 5 5 8 9 Mean: 1.31 cm Variance:.019 cm Standard Deviation:.139 cm Range: 1.13 cm Length (cm) Frequency 0.50-0.59 0 0.60-0.69 1 0.70-0.79 0 0.80-0.89 1 0.90-0.99 0 1.00-1.09 0 1.10-1.19 1 1.20-1.29 22 1.30-1.39 44 1.40-1.49 33 1.50-1.59 3 1.60-1.69 0 1.70-1.79 0 1.80-1.89 1 1.90-1.99 1
Univariate Data: Length Histogram and Modified Boxplot 5-Number Summary:.6, 1.3, 1.3, 1.4, 1.9 IQR:.1
Univariate Data: Length Pie charts
Univariate Data: Length Density Curves
Chapter 3 Examining Relationships
Bivariate Data: Length and Width Scatter Plot Explanatory Variable X: Length (cm) Response Variable Y: Width (cm) LSRL Equation: ŷ=.3652+.1853x r: 0.27267
Bivariate Data: Length and Width Residual Plot r-squared:.0743
Chapter 4 More about Relationships between Two Variables
Nonlinear Data: Length and Mass Nonlinear Scatter Plot Explanatory Variable X: Length (cm) Response Variable Y: Mass (g) LSRL Equation: ŷ=.0469+.0055x r: 0.27267 r-squared:.00016
Nonlinear Data: Length and Mass Transformed Scatter Plot Dimensional Transformation: Length Cubed LSRL Equation: ŷ=.8487+.0023x 3 r: 0.1899 r-squared:.00081
Nonlinear Data: Length and Mass Residual Plot of Transformed Data Two Influential Outliers: 5.8 cm and 6.8 cm
Categorical Relationships Two-Way Table Color of M & M M On Blue Yellow Red Orange Green Brown Total Yes 13 12 8 12 17 17 79 No 8 3 0 7 4 6 28 Total 21 15 8 19 21 23 107 Marginal Distributions: Yes: 79/107= 73.8% No: 28/107= 26.2% Blue: 21/107= 19.6% Yellow: 15/107= 14% Red: 8/107= 7.5% Orange: 19/107= 17.8% Green: 21/107= 19.6% Brown: 23/107= 21.5%
Chapter 5 Producing Data
Experimental Design Randomized Experiment Group 1: 50 Students Treatment 1: Taste Blue Treatment 1: Taste Blue 100 Students: Random Allocation Compare Results: Taste Preference Group 2: 50 Students Treatment 1: Taste Blue Treatment 1: Taste Blue
Chapter 6 Probability and Simulation
Probability Models Model One: Disjoint Event What is the probability than a flipped M & M will land M side up? S: { M side up, M side down} M up.56 M down.44
Probability Models Model Two: Non-Disjoint Event What is the probability than a randomly selected and flipped M & M will be orange and land M side up? S: {orange and M up, orange and M down, not orange and M up, not orange and M down} Orange.358 M up.082.105.455
Probability Models Two-Way Table M Side Blue Yellow Red Orange Green Brown Total Up 9 9 4 9 11 13 55 Down 12 6 4 10 10 10 52 Total 21 15 8 19 21 23 107
Chapter 7 Random Variables
Random Variables Discrete Random Variables Number of times a flipped M & M lands M side up
Random Variables Continuous Random Variables Distribution of mass
Chapter 8 Binomial and Geometric Models
Chapter 8 Binomial Questions Theoretical 107 M and M s are contained within a M and M packet. The probability that one will pick a red M and M at random is 0.07477. What is the probability of getting at least 3 red M and M s with replacement in 10 trials?
Binomial Distribution The mean of this data set is 0.7477 red M and M s. The standard deviation is 0.8317 reds away from the mean. The variance is 0.6917.
Chapter 8: Actual Binomial Trial # # of red M and M s picked In a sample size of 10 Probability of choosing a red M and M in trial 1 2 0.07477 2 1 0.07477 3 1 0.07477 4 2 0.07477 5 0 0.07477 6 0 0.07477 7 0 0.07477 8 0 0.07477 9 1 0.07477 10 1 0.07477
Geometric Questions 107 M and M s are contained within a M and M packet. The probability that one will pick a red M and M at random is 0.07477. What is the probability of picking the first red M and M at random on the 2th trial with replacement?
Actual Geometric Distribution Trial # Red M and M picked 2nd (1=yes, 0=no) 0 1 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 10 0 11 0 12 0 13 0 14 1 15 0 16 0 17 0 18 0 19 0 20 0 The actual probability of picking the first red M and M at random on the 2th trial with replacement is 0.05.
Chapter 9 Sampling Distributions
Chapter 9 Distribution of Colors in the population
Chapter 9 Distribution of Colors in the population
The mean of the sampling distribution of p-hat is p=0.07477 as. By the rule of thumb #1, the population size is greater than 10 times the sample size of 10 M and M s, so the standard deviation of p-hat is = 0.08317 Rule of thumb #2 does not apply in this situation because np>10 or n(1-p)>10, therefore a normal approximation may not be used. Due to the low frequency of red M and M s in our population, it would not be possible for rule of thumb # 2 to apply.
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