Homework 1 Solutions (1)

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Iowa State University *

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305

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Statistics

Date

Feb 20, 2024

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Uploaded by EarlCrown13355

Stat 305 Homework 1 Solutions 1.) Not graded, answers will vary. We just wanted you to think about some of the type of data you may come across in your engineering field. 2.) Not graded, answers will vary. The important idea was to think about the components of a two-factor experiment. As an example, suppose I wanted to learn about how different types of paper and different folds for a paper airplane affect the distance they fly when thrown. Key aspects are: Experimental Units : paper airplane templates Response Variable : Distance they fly when thrown Factors: Paper type and fold type Levels: For paper we may have two levels, copy paper, construction paper. For fold type, we may generically have three levels say fold A, fold B, and fold C. Treatments: We have six treatments here: Copy paper/Fold A, Copy paper/Fold B, Copy paper/Fold C, Construction paper/Fold A, Construction paper/Fold B, Construction paper/Fold B Principles of experimentation: We would use control by having the same person make the planes, the same person throwing them etc. We would want to do it in a controlled environment so wind doesn’t affect distance etc. I would randomly assign the experimental units to the treatments. I would want to have more than one experimental unit per treatment etc. 3.) Textbook Problems 1.1.4: Answers will vary. (10 pts total) Some examples of Concrete Populations: • All ISU undergraduate students • All registered voters in Iowa • All m&m’s in a five-pound bag Probability questions for these: • What is the probability a randomly chosen ISU undergrad is over 6 feet tall? • What is the probability of getting two republicans and two democrats in a random sample four voters? • What is the probability of getting at least three red m&m’s in a sample of 10 from the bag?

Statistical Inference questions for these: • What is the true average GPA of all ISU undergrad students? • What proportion of all registered voters are in favor of the new bill? • Is there evidence that the proportion of red m&m’s in the five-pound bag is greater than .25? Some examples of Conceptual Populations: • The battery lifetime of all Iphones • The speeds of cars on a particular highway • The time length of all MLB games Probability and Statistical Inference questions would be similar to the above 1.2.12: (6 pts total) a.) (34+27+15+5)/131 = 0.618 b.) (7+2+4+1)/131 = 0.107 c.) The histogram appears to be unimodal, centered around 4.5 – 5.0, slightly skewed to the right and no outliers present.

1.2.20: (5 points total) Both the stem and leaf plot look fairly uniform and skew off slightly to the right. Nothing really of interest stands out. From the histogram, a proportion of 23/47 = 0.489 are strictly less than 2000. A proportion of 17/47 = 0.361 are between 2000 and 4000. 1.2.22: Not Graded, just wanted to you interpret and comment on what the histogram can tell you. 1.2.29: (5 pts total). This question just asked for the frequency/relative frequency table and bar chart. Putting the data into JMP gives:

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1.3.34: (5 pts total) I first put the data into JMP to produce histograms and numeric summaries for both Urban homes (U) and Farm Homes (F) a.) I can get the sample means from the JMP output. The sample mean for Urban Homes is 21.54 and the sample mean from Farm Homes is 8.56. The average concentration (EU/mg) of settled dust is greatly higher in Urban homes. b.) The sample median for Urban Homes is 17 and the sample median for Farm Homes is 8.9. Again, the value for Urban homes is greater, but not as much as the difference in sample means. The sample mean is greater than the sample median value for Urban Homes because there is a very large outlier that will affect the mean by pulling it higher. 1.3.41: (6 pts total) a.) The sample proportion of successes is 7/10 = 0.70 b.) With coding S = 1 and F = 0, ࠵?̅ = %&%&’&%&%&%&’&’&%&% %’ = 0.70 . Same as above c.) We need +&, -. = .8 012345 6⎯⎯⎯8 ࠵? = 13 1.4.44: (5 pts total) Sample range = 182.6 – 180.3 = 2.3 .3 quantile: 13*.3 = 3.9 thus we interpolate between the 3 rd and 4 th ordered values. We get (.1)(180.5) + (.9)(180.9) = 180.86 Sample median = 181.6 (average of the 6 th and 7 th ordered values)

Sample mean = 181.408 Sample standard deviation = 0.724 1.4.46: (10 pts total) a.) The distributions for Control and Warmer are both skewed left and there are outliers in each distribution so I will use the median as my value of center (Note the mean would also be fine here). Control: median = 1.9 Cooler: median = 1.76 Warmer: median = 2.305 The centers are fairly similar with the Warmer being the largest and Cooler being the smallest. The sample means are also similar. b.) Standard Deviations: Control: s = 0.531 Cooler: s = 0.401 Warmer: s = 0.778 The interpretation for all of these is the “average” distance observations lie from their respective means. The sample standard deviations are similar with Warmer having the biggest spread given by the standard deviation. c.) IQR’s (or “fourth spreads” as the book puts it) Control: IQR = 0.51 Cooler: IQR = 0.47 Warmer: IQR = 0.785 The IQR’s for each group are nearly equal to their standard deviations and they both give us a good representation of the variability in the data. 1.4.57 (5 pts total) a.) The standard 1.5IQR cutoffs are 164.8 and 248 The 3IQR cutoffs for “extreme” outliers are 133.6 and 279.2 Based on these, 125.8 is an “extreme” outlier and 250.2 is a standard outlier.

b.) Nothing really interesting with the boxplot. Pretty standard 1.4.61: Not graded , just wanted you to comment on what similarities/differences you see in the boxplots

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