Solution1_Winter 2024 V6 PDF

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Stat 151-Assignment 1 STAT 151 Assignment 1 Due date : refer to the Course Outline Purposes This assignment has two parts. The first part assesses your knowledge of sampling, population, descriptive vs. inferential study, and designed experiment vs. observational study. The first part also assesses your understanding of choosing proper numerical and graphical tools to summarize data. The second part assesses your skills in using R commander to create graphs and interpret their meaning and obtain descriptive measures of a given data set. Instructions Important note : For every assignment in this course, you are required to complete the questions or tasks in Part A by hand. This means that to do any calculation or drawing, you will NOT use R commander or any computer application. That is, you are meant to do the calculations manually with a non- programmable scientific calculator and use a pen or pencil to draw figures or build a distribution table on paper (or on an iPad/tablet). Then you will submit a photo of your written solution using the appropriate submission box on the corresponding Crowdmark submission page. Before you complete Part B using R commander, you should read and practice the R commander steps by following the related examples in the Lab Manual and the Demos, which you can download via a link in the Course Content folder on Moodle. Graphs should be labelled appropriately, and units should be included in answers when appropriate. Final answers should be rounded to 3 decimal places. Part A-Lecture Component 1. A researcher is studying how the seriousness of local current issues described statistically in introductory videos impacts the perceived interest expressed by students for learning Statistics at MacEwan. The researcher obtains a random sample of 1000 students and randomly assigns them to 4 groups. Students in group 1 saw a video that statistically described serious current local issues, students in group 2 saw a video that statistically described uplifting current local issues, students in group 3 saw a video that statistically described a mix of serious and uplifting current local issues, and students in group 4 did not view a video. Following this, students were asked whether they were interested in learning Statistics.

Stat 151-Assignment 1 1 After completing her research, the researcher noted that 70% of the students in group 1 expressed interest in taking the course, 74% of the students in group 2 expressed interest in taking the course, 78% of the students in group 3 expressed interest in taking the course, and 66% of the students in group 4 expressed interest in taking the course. Based off this, the researcher stated that watching videos that statistically describe local current issues seems to positively impact the perceived level of interest in taking Statistics by MacEwan students, and that providing statistical description about a mix of serious and lighthearted local current issues seems to invoke the most interest. a) Identify the sample and population of this study. (2 marks) b) Is this study a designed experiment or an observational study? Explain why. (2 marks) c) Is this study descriptive or inferential? Explain why. (2 marks) d) Propose two different methods to randomly assign the 1000 students to each of the 4 groups. Hint: think of this as obtaining a simple random sample of 250 students from a collection of 1000 students, then taking a second random sample of 250 students from the remaining 750 students, and so on, in order to obtain 4 separate random samples, each of size 250. (2 marks) a) The population is all students at MacEwan. The sample is the 1000 students in the study. b) Students are randomly assigned to the 4 treatment groups. This is a designed experiment. c) This study is inferential. The researcher makes a statement about all students at MacEwan using results obtained from a sample of only 1000 MacEwan students. d)Possible answers include: (1). Order students by last name and assign to each student a number between 1 and 1000. Write numbers from 1 to 1000 on paper slips and put them in a box. Shake the box and take out 250 paper slips; the students whose last names correspond to the first 250 numbers are placed in group 1, the students whose last names correspond to the second 200 numbers are placed in group 2, the students whose last names correspond to the third 200 numbers are placed in group 3, and the remaining 250 students are in group 4. (2). Order students by last name and assign to each student a number between 1 and 1000. Use a random number table. We need four columns to represent 1-1000. Move through the table until you obtain 1000 unique values between 1-1000: the students whose last names correspond to the first 250 numbers are placed in group 1, the students whose last names correspond to the second 250 numbers are placed in group 2, the students whose last names correspond to the third 250 numbers are placed in group 3, and the remaining 250 students are in group 4. (3). Use software or an online website to randomly permute the numbers 1-1000. The students whose last names correspond to the first 250 numbers are placed in group 1, the students whose last names correspond to the second 250 numbers are placed in group 2, the students whose last names correspond to the third 250 numbers are placed in group 3, and the remaining 250 students are in group 4.

Stat 151-Assignment 1 2 2. The city of Edmonton retains a current database of assessed dwelling values for all addresses in Edmonton. For the neighbourhood of Highlands, the data below provides a random sample of assessed dwelling values for 15 households. The values (in thousands) are below. 555, 413, 426, 427, 432, 72, 258, 316, 336, 384, 448, 453, 525, 549, 391 Complete the following questions. Round your final answers to the third decimal place. a) Draw a histogram based on the data. Use 0, 100, 200, 300, 400, 500 and 600 as your cutpoints. What can you tell about the distribution of the data? (3 marks) The data is left skewed. There is one household assessed at only 72K that pulls the tail of the distribution left. The center appears to be about 400 and the bulk of the data is spread between 300 and 500. Grader: please note if the student has an earlier version of the

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Stat 151-Assignment 1 3 assignment without specified cutpoints, full marks are to be awarded for a reasonable histogram. Please refer to the following table for calculation of several of the answers in parts below. RANK X X - 399 (X-399) 2 MIN = 72 1 72 -327 106929 2 258 -141 19881 3 316 -83 6889 POSITION (15+1)/4 = 4, Q1 = 336 4 336 -63 3969 5 384 -15 225 6 391 -8 64 7 413 14 196 POSITION (15+1)/2 =8, MEDIAN = 426 8 426 27 729 9 427 28 784 10 432 33 1089 11 448 49 2401 POSITION 3(15+1)/4 = 12, Q3 = 453 12 453 54 2916 13 525 126 15876 14 549 150 22500 MAX =555 15 555 156 24336 TOTAL 5985 0 208784 MEAN = 5985/15 399 VAR = 208784/14 14913.14286 STDEV = SQRT(14913.14286) 122.1193795 RANGE = MAX - MIN = 555 - 72 = 483 IQR = 453-336 = 117 NO MODE b) Find the mean, median, and mode of the data. (6 marks) Mean = 399K (see table above for calculations) Median = 426K (the 8th observed value, the middle value in the observed data), No mode c) Calculate the standard deviation of the data. (5 marks) 122.119K to 3 decimal places (see table above) d) Find the range of the data. (2 marks)

Stat 151-Assignment 1 4 Max – Min = 555 – 72 = 483K (see table above) e) Draw a box plot based on the data. What can you tell about the distribution of the data? (5 marks) There is one outlier, observation 13 (value is 72K). The middle 50% bulk of the data lies between about 370K and 450k. The data is left skewed, as the median (about 430K) is high in the box, and there is a low outlier. Q1 is around 370K, Q3 is around 450K. The max is around 570K, and the min is around 75K. f) Obtain and interpret the five-number summary. (10 marks) (There are many approaches to calculating quartiles. R actually offers 9 options. The answers to two approaches are below. Students may use any sensible method for finding quartiles. Graders should be aware that instructors may use different methods to teach this topic. Simply watch for consistency in student answers in f, g and h.) Table Approach (using Quartiles from Table Above): Min = 72 , Q1 =336 , Median = 426 , Q3 =453 , Max = 555 (all in thousands) Interpretation with Table Approach for Quartiles: In thousands, 25% of the data lies between 72K and 336K, 25% of the data lies between 336K and 426K, 25% of the data lies between 425K and 453K, and 25% of the data lies between 453K and 555K. The distance between Q1 and the median is 90K, while the distance between the median and Q3 is 28K (with the second distance being considerably smaller). The value of 72K is an

Stat 151-Assignment 1 5 outlier. It lies quite a bit below the next highest assessed data value of 226K (from the raw data). Textbook Method for calculating quartiles: There are 15 observations. The median Q2 is the value in the middle of the sorted list (72, 258, 316, 336, 384, 391, 413, 426, 427, 432, 448, 453, 525, 549, 555). The first half is (72, 258, 316, 336, 384, 391, 413, 426) and the second half is (426, 427, 432, 448, 453, 525, 549, 555). The first quartile is the median of the first half (first 8 observations): Q1 = (336+384)/2=360. The third quartile is the median of the second half (last 8 observations): Q3=(448+453)/2=450.5. 5 number summary is (min, Q1, Q2, Q3, max) = (72, 360, 426, 450.5, 555) Interpretation with Textbook Approach for Quartiles: In thousands, 25% of the data lies between 72K and 360K, 25% of the data lies between 360K and 426K, 25% of the data lies between 426K and 450.5K, and 25% of the data lies between 450.5 K and 555K. The distance between Q1 and the median is 66K, while the distance between the median and Q3 is 24.5K (with the second distance being considerably smaller). The value of 72K is an outlier. It lies quite a bit below the next highest assessed data value of 226K (from the raw data). g) Obtain and interpret the interquartile range. (4 marks) Table Method: IQR = Q3 – Q1 = 453 - 336 = 117K Textbook Method: IQR = Q3 – Q1 = 450.5 - 360 = 90.5K Graders: If student used another method, they should use the Q1 and Q3 they obtained in part f. h) Are there any potential outliers? Justify by calculation. (3 marks) With quartiles from Table Approach: Q1 – 1.5*IQR = 336 – 1.5(117) = 336 – 175.5 = 160.5K Q3 + 1.5*IQR = 453 + 1.5(117) = 453 + 175.5 = 628.5K 72K is identified as an outlier as it is below 160.5K. There are no identified outliers above 628K. With quartiles from Textbook Approach: Q1 – 1.5*IQR = 360 – 1.5(90.5) = 360 – 135.75 = 224.25K Q3 + 1.5*IQR = 450.5+ 1.5(90.5) = 450.5 + 135.75 = 586.25K 72K is identified as an outlier as it is below 224.25K. There are no identified outliers above 586.25K. Graders: If student used another method, they should use the Q1 and Q3 they obtained in part f.

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Stat 151-Assignment 1 6 i) Choose the most appropriate measures of center and spread for this data. Justify your choice. (3 marks) As the data is left skewed, the median and the IQR are the most appropriate measures of center and spread for this data. j) Do you find either a histogram or a boxplot preferable in describing the shape of data when a sample size is 15? Explain your reasoning. Were both useful in this problem? (3 marks) Both are useful with a data size of 15. A histogram will have a small number of bins and a small number of values for the bins, but should give a general indication of center, spread and shape. Both a histogram and a boxplot will indicate if data is skewed, although a boxplot cannot identify if data is normal or multi-peaked. Both a histogram and a boxplot can indicate outliers. For this data, both the histogram and the boxplot identified left skewness and a left outlier. For this small sample of skewed data, the histogram did not tell us any more about the shape of the data than the boxplot. k) Would you prefer using a boxplot or a histogram if the sample size was only 8? Why? (3 marks ) If the sample size was only 8, a histogram would likely not have enough bins or enough values in any bins to be particularly useful in describing the shape of the distribution. The boxplot would help a bit more as it would provide a 5 number summary with an IQR and it would more readily identify any outliers. Part B-Lab Component Finish the following questions using R and R commander. 1. The survey data STATISTICSSTUDENTSSURVEYFORR contains several columns, including the columns NUMVINYL (number of vinyl records a student owns), WKHRSHWK (weekly hours of homework per course by a student), LIKEENGLISH (how much a student likes the subject English on a scale of 1 (dislike very much) to 5 (like very much)) , IMPISSUE (most important election issue a student feels faces us today, including criminal, domestic policy, education, electoral, environmental, foreign policy, healthcare, immigration, and social). For each of these 4 columns, indicate if the variable is qualitative (non-ordinal), qualitative (ordinal), quantitative (discrete) or quantitative (continuous). (4 marks) NUMVINYL (number of vinyl records a student owns) – quantitative discrete WKHRSHWK (weekly hours of homework per course by a student) – quantitative continuous LIKEENGLISH (how much a student likes the subject English on a scale of 1 (dislike very much) to 5 (like very much)) – qualitative ordinal IMPISSUE (most important election issue a student feels faces us today, including criminal, domestic policy, education, electoral, environmental, foreign policy, healthcare, immigration, and social) – qualitative non-ordinal

Stat 151-Assignment 1 7 2.The data file HIGHLANDS contains the assessed dwelling values of ALL residential Edmonton addresses in the HIGHLANDS. ( FYI: The file HIGHLANDS is a subset of https://data.edmonton.ca/City-Administration/Property-Assessment-Data-Current-Calendar- Year-/q7d6-ambg ) a) Create and paste a histogram that summarizes the assessed residential dwelling values for the HIGHLANDS dwellings. You will need to make 20 bins to get a meaningful picture. Comment on the distribution of the assessed dwelling values in terms of overall shape, modality, symmetricity/skewness if applicable. (4 marks) The distribution is right skewed. There are a few expensive houses in the neighbourhood (mostly on Ada Blvd if the student looks it up) assessed at considerably more than other households in the neighbourhood that pull the tail of the distribution right. (Students can make any sensible number of bins, but an automated number is too small). Roughly 70% of the houses are assessed at less than roughly 600K, and the median looks to be around 400K. The average house price will be pulled up by the households costing between about 1 million to 3.5 million. b) Create and paste a boxplot that summarizes the assessed residential dwelling values for the HIGHLANDS dwellings. What can you tell about the distribution of the data? (4 marks) The bulk of the data, between Q1 and Q3, falls between about 350K to 500K (very roughly, from the boxplot). There is a short tail to the left that shows some less expensive properties were assessed and a much larger tail to the right that shows several expensive properties were

Stat 151-Assignment 1 8 assessed, including one extreme outlier at 3.5 million. In this case, the mean will be pulled above the median as there are many more outliers to the right than the left. c)Find and paste a full set of descriptive measures for the entire population of assessed HIGHLANDS residential dwelling values. Choose the most appropriate descriptive measures and explain your choice (with reference to the shape of the data distribution). (4 marks) As the data is right skewed, the median of 401K is the most appropriate measure of centrality, and the IQR of 520K-340K = 162K is the most appropriate measure of spread. 50% of the assessed home values will fall between 340K and 502K. The distance from the minimum to Q1 is quite notable (about 338K), as is the very much larger distance of about 3500K from Q3 to the maximum. The distance from Q1 to the median is about 60K, while the distance from the median to Q3 is about 101K. The mean of 440K is pulled above the median of 401K by the values of the outlying highly assessed homes. Overall, if one ignores the outlying values, one can likely get a fairly nice home for around 400K in this neighbourhood. d) You will notice that the histogram distribution of assessed residential dwelling values that you found for the Highlands in the sample of size 15 taken in Part A does not match the shape of the distribution you found when you used the HIGHLANDS datafile with the assessed dwelling values for all residential dwellings in the Highlands in Part B. Furthermore, the boxplot of assessed residential dwelling values that you found for the Highlands in the sample of size 15 taken in Part A does not match the shape of the boxplot you found when you used the

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Stat 151-Assignment 1 9 HIGHLANDS datafile with the assessed dwelling values for all residential dwellings in the Highlands. State how the shapes differ and explain why this may have happened. (4 marks) The sample did not obtain any of the very high values in the dataset, but did obtain a lower value from the dataset. Thus, the small sample (size 15) ended up with a left skewed distribution in spite of the actual population of all assessed values being right skewed!! Even though it was indicated that the sample was taken randomly (as appropriate), the random sample ended up being nonrepresentative. 3.The data set HEARTFAILUREPREDICTION looks at several variables that play a role in heart failure prediction for 918 people in the United States. ( For the curious, this set of open data can be found at https://www.kaggle.com/datasets/fedesoriano/heart-failure-prediction .) a)For each of the numerical variables Age (in years) and MaxHr (maximum heart rate achieved in bpm), find and paste two suitable graphs to summarize their distribution. Comment on each distribution in terms of overall shape, modality, symmetricity/skewness if applicable. (6 marks) The distribution of age is close to normal. There are more people above 50 years of age than below 50. The data appears roughly balanced around a center of 55 years. The younger folk may pull the mean a bit below from the median of about 55. The data is nicely spread out and we can readily use standard deviation as a measure of spread, as it certainly appears that the bulk of the data falls within 3 standard deviations of the mean and one would expect with a normal shaped distribution.

Stat 151-Assignment 1 10 This data is also relatively normal shaped, although the bell does not taper quite as evenly as it does for a normal distribution. The left tail is slightly longer with a mere suggestion of a skew, and the boxplot identifies two outliers on the left. The data appears to be centered around the value of 130 (both median and mean). Again, we can likely feel comfortable using standard deviation as our measure of spread, even though we do identify two slight outliers. b)For each of the numerical variables Age and MaxHr, find and paste a full set of descriptive measures. Choose the most appropriate descriptive measures in each case and justify your choices (with reference to the shapes of their distributions). (6 marks) For Age, the best measures of center and spread are mean and standard deviation. The age data is very close to normal. For MaxHr, it seems that the best measure of center and spread are again mean and standard deviation as the data is close to normal. However, in this case, the boxplot and histogram both show a slight tendency to a left skew in the data. In this case, it is a bit helpful to also consider the median, range and IQR. The mean of the MaxHr distribution is pulled down a bit from the median. c) Find and paste the most suitable graph to show the relationship between Age and MaxHr (maximum heart rate) among the 918 respondents in this dataset. Briefly describe the relationship you see. (4 marks)

Stat 151-Assignment 1 11 As these are two numerical variables that have a “relationship”, a scatterplot is the most appropriate graph. As age increases, the maximum heart rate attainable decreases. It is possible that a line could be fit to the data that would offer a good model of prediction of MaxHr for a given Age. 4.The dataset HENSANDBEES summarizes the locations in Edmonton where residents are housing hens or keeping bees. (FYI, t his dataset was taken from https://data.edmonton.ca/Community-Services/Hens-and-Bees/trz2-qkzs ) a)Find and paste a suitable graph that summarizes how the counts in the data in the numerical column that indicates the number of hens for households that were granted a hen license is distributed. Look in the R dropdown of graphs and explore to find a graph that you think best accommodates the fact that there is a low number of distinct potential numerical values for #hens (2,3,4,5,6,7,8). Be sure to explain why you chose this graph. You will want to think about this one and suggest something perhaps not covered in lab! (4 marks) ( Either the stem and leaf or the discrete values graph are most acceptable here for full marks. I suggest part marks for other choices such as a histogram (lumps into too few bins) or a boxplot (less tail information, misses dip at 5) or a dotplot (no counts )) . Instructors should feel free to direct their graders as to how to grade this question based on how they teach their students. The discrete values graph is most useful here as the “number of hens” variable is disrete and takes on only the values 2,3,4,5,6,7 and 8. The discrete value graph emphasizes that your data is not continuous as you have a count spike. A stem and leaf plot is also most useful here for the same reasons. It also provides counts, but it should be made so your stems are discrete.

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Stat 151-Assignment 1 12 COUNTS OF EDMONTON LICENSED HOUSEHOLDS BY NUMBER OF HENS AT HOUSEHOLD Note: stem and leaf does not allow titles so the title was typed in afterwards. 5a): Draw side by side histograms to compare the maximum heartrates of male and female respondents in the HEARTFAILUREPREDICTION dataset. Decide if it is more appropriate to use counts or relative frequencies (percents) on the histogram. Indicate why you made the choice you did. Describe your chosen graphs. What further graphs do you think you might consider to be of interest after looking at these two histograms? (7 marks)

Stat 151-Assignment 1 13 Why choose percentage: There are almost 4 times as many males (725) as females (198) in the study (see numerical summaries below). Percentages make for histograms that are more appropriately comparable in a visual and descriptive explanatory sense. Description of percent histograms: For the females, the data is slightly left skewed, but appears somewhat normal. For the males, the date is more balanced with two tails of about the same length. It appears that men were more likely, percent-wise to attain a lower maximum heart rate, while females were more likely, percent wise, to attain a higher maximum heart rate. The female center looks to be about 150bpm and the male center about 130bpm. The female graph looks to have a range of up to about 120bpm, while the male graph is more spread out with a range of up to about 150bpm. In both cases, as the graphs are “relatively” normal, standard deviation is a good measure of spread. (From part b, standard deviation is about 22 bpm for the females and 26 bpm for the males). Suggestions for further graphs: What these graphs don’t tell us is how the data was distributed age wise for the females and males in the data set. Looking at scatterplots of Age and MaxHr for the males and females separately would certainly be of interest. (Students may have other interesting and valuable answers here – all that matters is that they think a bit here 😊 ) 5b): Find proper numerical summaries to compare the maximum heartrates of male and female respondents in the heart dataset, remembering to base your choices on the distribution shapes obtained above. Briefly explain how your numerical summaries elucidate your graphs.(5 marks) The mean and median for MaxHr are quite close for the males (about 134bpm) and also quite close for the females (146bpm and 150bpm), as we would expect from the data graphs for the two sexes being fairly normal in both cases. The center of the data is higher in bpm for the females than the males. The spread (range and standard deviation) of data is larger for the males (142bpm and 25.719 bpm) than the females (102 bpm and 22.155 bpm) as can be seen in the numerical summaries and on the graphs. Some of the males appear to have quite low maximum heart rates. One could investigate if these were older males with a scatterplot of Age versus MaxHR, as mentioned above. As the data is “close” to normal is both graphs, mean and standard deviation are appropriate measures of center and spread. 5c: Consider the variable “MaxHr” in the heart dataset. i) Calculate the interquartile range of the variable “MaxHr” for females. (2 marks) IQR = 163 – 130 = 33 BPM ii) What percent of distribution data falls within the interquartile range of the “MaxHr” variable for the females? (1 mark) 50% iii) Calculate the interquartile range of the variable “MaxHr” for males. (2 marks) IQR = 152-117 = 35 BPM

Stat 151-Assignment 1 14 iv) What percent of distribution data falls within the interquartile range of the “MaxHr” variable for the males. (1 mark) 50% Submission Submit your work by accessing the Crowdmark email (or Crowdmark link on Moodle) to submit Assignment 1. Please ensure that each picture is properly oriented and easy to read (not fuzzy, not too small, and not taken in a dark room so that it is difficult to read). All work must be submitted to Crowdmark by 6:00 PM on the due date. Avoiding Plagiarism: If you submit an assignment, you are claiming it is your work. Do not allow any part of your work to be copied by anyone else. Where two or more assignments are found to be unreasonably similar, either in whole or in part, and no assistance has been acknowledged, all parties involved are liable to a score of zero on the assignment. MacEwan University’s academic policies are available at: https://www.macewan.ca/contribute/groups/public/documents/policy/zwdf/cg9s/~edisp/student_ acad_integ_policy.pdf

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