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HSCI 190 Fall 2023 HSCI 190 – Module 1 Homework Problem #1 A past study investigated the bloodwork of eight individuals who experienced the unusual occurrence of vitamin D intoxication. It was suspected the vitamin D intoxication might have been caused by excessive dairy milk consumption. To confirm this suspicion, patients’ calcium and phosphorus levels were collected. In a healthy adult: Normal range for calcium levels = 2.12 to 2.74 mmol/l Normal range for phosphorus levels = 0.74 to 1.45 mmol/l The data from the eight patients is presented below: Patient Calcium (mmol/l) Phosphorus (mmol/l) 1 2.92 0.94 2 3.84 1.91 3 2.37 1.23 4 2.99 1.42 5 2.67 1.03 6 3.17 1.10 7 3.74 1.52 8 3.44 1.42 A. Identify the level of measurement of calcium and phosphorus in this data set and explain your reasoning. (2 marks) The level of measurement of calcium and phosphorus in this data set would be scale data because the table shows numerical measurements. More specifically, the type of scale data would be ratio since there is a real zero (the absence of any calcium and/or phosphorus). B. Identify the mean, median, mode, standard deviation, range, and interquartile range of the recorded calcium levels and phosphorus levels. (12 marks) Calcium Phosphorus Mean (mmol/l) 3.14 1.32 Median (mmol/l) 3.08 1.33 1

HSCI 190 Fall 2023 Mode (mmol/l) N/A 1.42 Standard Deviation (mmol/l) 0.51 0.32 Range (mmol/l) 1.47 0.97 IQR Exclusive (mmol/l) 0.93 0.45 C. Write a summary of your findings and whether you believe the patients fall within the normal calcium and phosphorus ranges based on your findings. (2 marks) The data set was of the calcium and phosphorus levels of eight individuals who experienced the unusual occurrence of vitamin D intoxication. The average/mean calcium level in this study was 3.14 mmol/l, with a standard deviation of 0.51 mmol/l. Based on the mean, the patients do not fall within the normal calcium ranges because the mean was calculated to be 3.14mmol/l which is not in the normal range of 2.12 to 2.74 mmol/l. The standard deviation tells us that most of the patients falls within 0.51mmol/l of the mean (2.63mmol/l to 3.65mmol/l), although the standard deviation is small, it is still higher than normal calcium levels when looking at its variance from the mean. The median calcium level in this study was 3.08, with 50% of the observations being within 0.93 mmol/l of the median. The range between the maximum and minimum calcium levels is 1.47 mmol/l. Then mode is not applicable in this data set because there are no repeated values. The average/mean phosphorus level in the study was 1.32 mmol/l, with a standard deviation of 0.32 mmol/l. Based on the mean, the patients do fall within the normal phosphorus ranges because the mean was calculated to be 1.32mmol/l which is within the normal range of 0.74 to 1.45 mmol/l. Additionally, the standard deviation is a fairly small number meaning that there isn’t much variance around the mean. Based on the mean and standard deviation we can conclude that most of the patients’ phosphorus levels are close to the mean, which is in the normal range. The median phosphorus level in the study was 1.33, with 50% of the observations being within 0.45 mmol/l of the median. The range is 0.97 mmol/l and the mode is 1.42 mmol/l. Problem adapted from: Jacobus, C.H., Holick, M.F., Shao, Q., Chen, T.C., Holm, I.A., Kolodny, J.M., Fuleihan, G.E.H., and Seely, E.W. (1992). Hypervitaminosis D Associated with Drinking Milk. The New England Journal of Medicine, Vol 326, 1171- 1177. Problem #2 The Western Ontario and McMaster (WOMAC) Universities Osteoarthritis Index is widely used to assess pain, stiffness, and function in patients with osteoarthritis of the hip or knee. An excerpt of the scale is shown below: 2

HSCI 190 Fall 2023 A. Describe the level of measurement this question uses, and explain your rationale (i.e., why you believe that level of measurement is the correct choice). (1 mark) I believe that the order of measurement for this question would be categorical data and more specifically ordinal data since it creates a set of categories that are ordered/ranked in terms of the severity of pain. Because each category holds a different magnitude, this would be considered ordinal data. B. Imagine you collected the following data, calculate one measure of central tendency and one measure of variance to summarize this data. Provide your justification for your selected choices. (3 marks) *table typo updated Sept 11/23 - rating scores (I.e., numbers) did not change, only the names for the category (I.e., severe, etc.) Participant 1 2 3 4 5 6 7 8 9 Rating A None Mild Severe Severe Severe None Mild Mod. Extreme Rating Score (0-4) 0 1 3 3 3 0 1 2 4 Mode – 3 Range – 0 to 4 This data set shows the pain index of 9 individuals on a scale of 0-4 (0-none, 1-mild, 2-moderate, 3-severe, 4-ratio). I chose to use the mode as the measure of central tendency and range as the measure of variance. The mean in this case is not suitable for ordinal data because we are looking at a scale of 0-4, information is better represented using mode because it tells us what the most 3

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HSCI 190 Fall 2023 common answer is on that scale of 0-4. Using mean does not give us a selectable answer which fails to make sense when summarizing this information. The variance I chose to use is range because it shows variance from mode. Range in this case is 0 to 4 and since we are using mode, we can state what the most common choice is on the scale of 0 to 4. Pairing mode with range makes sense when explaining the data. Problem #3 Consider the following data set: Blood O B A AB AB O O A B O B Types A B A A O AB B A O B AB B O O A AB A AB A B A O A. Arrange the data in as an absolute frequency table. (1 mark) Bood Type Number of Participants A 10 B 8 AB 6 O 9 Total 33 B. What is the relative frequency of individuals with type B blood? (1 mark) Frequency of B = 8 Total = 33 Relative Frequency = Absolute Frequency ∑ of allfrequencies ¿ 8 33 = 0.2424 ∙ 100 = 24.24% 4

HSCI 190 Fall 2023 The relative frequency of individuals with type B blood is 24.24% 5

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