MAS 183 2017 final Exam Answers

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Apr 3, 2024

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OFFICIAL Answers to Practise Test Q 1 Volume counted = 9 mm 2 x 0.1 mm = 0.9 mm 3 = 0.9 uL. a. What is the cell count per microlitre (µL)? = 20 x 17 per 0.9 uL = 340 cells /0.9 uL = 378 cells /uL b. What is the cell count per millilitre (mL)? = 378 x 10 3 / mL or 3.78 x 10 5 cells / mL c. What is the cell count per Litre? 3.78 x 10 8 cells / L
OFFICIAL Answers to Practise Test Q 2
OFFICIAL Answers to Practise Test Q 3 correlation coefficient = 0.63 As the correlation coefficient (r=0.63) is significantly larger than 0.3, I conclude that there is a significant positive correlation between the counts in chambers 1&2 of each slide. Note: Using XLSTAT, there is significant evidence of positive correlation (p value = 5.4E-5 200 300 400 500 600 700 800 300 350 400 450 500 550 600 650 700 R² = 0.39 Count (Chamber 1) Count (Chamber 2)
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OFFICIAL Answers to Practise Test Q 4 3 0 1 3 1 4 7 3 2 2 3 7 3 3 0 2 8 9 3 4 1 2 6 9 3 5 1 1 1 2 2 2 3 4 3 6 2 5 6 8 3 7 0 5 7 7 8 3 8 0 5 5 3 9 3 8 4 0 4 7 9 4 1 4 2 4 3 9 Q 5 Median = Q 6 Provide a 5 number summary of the data: Min = 301 Q 1 = 340.5 (Using Quartile.inc)
OFFICIAL Answers to Practise Test Q 2 = 352.5 Q 3 = 377.25 Max = 439
OFFICIAL Answers to Practise Test Q 7 250 270 290 310 330 350 370 390 410 430 450 Diameter (µm) Q 8 Lower Limit = 340.5 – 1.5*36.75 = 285 Upper Limit = 377.25 + 1.5 * 36.75 = 432 ( IQR = 377.25 – 340.5 = 36.75) Does all the data lie between the lower and upper limits? NO, 439 is an outlier Discuss briefly, the action you would take if you discovered an outlier in the data set. Note that these are the inner fences; therefore it is not unusual to find a value lying outside. Check with supervisor or colleague as necessary. Proceed with caution as these “points of interest” may belong inside the population. If the outer fences are calculated (Q 3,1 ± 3*IQR)– data points lying outside these are highly suspect and should probably/definitely be removed from the data set. In this case we are advised it should be removed from the data set. DATA POINT NOTED DOWN, THEN ERASED: (439)
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OFFICIAL Answers to Practise Test Q 9 Data point 439 removed from the data set Diameter ( μm ) Midpoint x i Tally Frequency 301-320 310 III 3 321-340 330 IIII II 7 341-360 350 IIII IIII II 12 361-380 370 IIII IIII 10 381-400 390 IIII 4 401-420 410 III 3 Q 10 On the graph paper provided, draw a fully labelled frequency histogram for the data. 310 330 350 370 390 410 430 0 2 4 6 8 10 12 14 Diameter (µm) Frequency Q 11 Average = 357 (3sf) Standard deviation = 26.8 (3sf) Q 12 CV(%) = 7.5 %
OFFICIAL Answers to Practise Test Q 13 x t s n < μ < x + t s n x = 357.3 s = 26.80 n = 39 t = 2.024 357.3 2.024 26.8 39 < μ < 357.3 + 2.024 26.8 39 =348.6 < u < 366.0 µm (4sf) Box and Whisker Plot 329 367 324 319 379 344 371 363 360 356 300 320 340 360 380 400 ppm (Cu) Q 14 BLUE (bottom) * Larger IQR , Q3 and Max * Smaller Min, Q1 and Q2 * has less symmetry
OFFICIAL Answers to Practise Test Which of the two data sets described by the Box and Whisker Plot (above) is more likely to follow a normal distribution – give at least one reason for your answer. Q 15 Both data sets can be modelled using a normal distribution as they display strong central tendency. However of the two, the top data set: * is more symmetrical * The range is smaller suggesting more precision in the measurements Q 16 Any one of these …… Central Tendency Symmetrical distribution 68 / 95 / 99.7% rule Median = Mode = Mean Asymptotic to x axis Q 17 (b) Use the statistical tables to find P(z -1.25) P(z -1.25) = 1 - P(z < -1.25) = 1 – 0.0401 = 0.9599 Q 18
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OFFICIAL Answers to Practise Test Q 19 Use the statistical tables to find P(-0.87 z 1.26) P(-0.87 z 1.26) = P(z 1.26) - P( z -0.87) = 0.8962 - 0.1922 = 0.7040 Q 20 Find the value of z (below). z = 1.96 Q 21 Measurement Measurement Uncertainty % Uncertainty Length (mm) (L) 4.51 0.02 0.44% Width (mm) (W) 5.50 0.02 0.36% Depth (mm) (D) 12.6 0.02 0.16% Volume (LDW) 313 (3sf) ±1.9 cm 3 0.60% Calculate Right column results first. Calculate Volume = LDW next. Then calculate the measurement uncertainty on the volume as 0.6% of 313