Question

Instructions Read each scenario in the document below and answer the questions. When you have completed all scenarios, save the document with your last name in the file (ex: Hoernemann Module 13 Quality Control Assignment) and submit as an attachment to the appropriate assignment folder. Some scenarios require you to perform quality control calculations – you may use an Excel spreadsheet to perform the math but must include the spreadsheets with your submission for this assignment to receive full credit! Scenario 1 You are the chief technologist for a clinical chemistry lab. You have a new batch (lot number) of control, and you wish to know whether you should use it. You assay it and obtain the following QC data for the sodium analyses on Wonder Control level 1: 145 135 143 146 151 136 137 133 148 142 134 140 138 146 141 138 140 138 139 142 1. Determine the mean, median, and mode 2. Decide whether these results demonstrate a normal (Gaussian) distribution or not. How do you know? 3. Based on the mean/median/mode, and distribution, would you use this control? Why or why not? Scenario 2 You are the chief technologist of a clinical chemistry laboratory. You must choose between two different glucose test methods for use in the lab. You assay a specific normal glucose control by both methods and obtain the following results: Glucose Method A Glucose Method B 100 105 101 99 107 100 93 99 102 98 104 96 110 105 91 94 101 97 99 103 89 108 111 92 100 98 102 101 102 98 96 104 99 95 100 100 95 97 105 103 1. For each method, determine the mean, median, mode, and SD. 2. Are both methods showing a normal distribution? 3. Which method is more precise? How do you know? 4. The manufacturer has listed a mean of 100 mg/dL and an SD of + 3 mg/dL. Which method is more accurate? How do you know? 5. Which method would you choose for your lab and why?

Answer 1

Scenario 1: Extract the data in table form as shown below: S. No. Control level 1 145 135 143 146 151 136 137 133 148 142 10 11 12 13 14 15 134 140 138 146 141 138 140 138 139 142 16 17 18 19 20

Calculate the mean, median, and mode using MS excel spreadsheet as shown below: Formulate MS excel spreadsheet as shown below: А B Control level 1 21 4 5 3 4 6 5 145 135 143 146 1151 136 137 133 148 76 142 9 8 10 9 11 10 12 11 13 12 14 13 15 14 16 15 17 16 18 17 19 18 20 19 21 20 22 134 140 138 146 1141 138 140 | 138 139 142 Mean =AVERAGE(B2:B21) Median =MEDIAN(B2:B21) Mode =MODE(B2:B21) Figure 1: Formulated spreadsheet 23

The output of above formulation is shown below: А S. No. 1 2 3 5 6 7 8 Control level 1 145 135 143 146 151 136 137 133 148 142 134 140 138 146 9 10 11 12 13 15 14 15 141 16 17 18 19 21 16 138 17 140 18 138 19 139 20 142 Mean 140.60 Median 140 Mode 138 Figure 2: Final output 24

Hence, the mean is 140.6, median is 140 and mode is 138) for control level 1. The mean, median, and mode are not equal. Therefore, it is not a normal distribution. Calculate the distribution as shown below:

S. No. 21 3 2 4 3 5 4 65 76 8 7 98 Control level 1 145 135 143 146 151 136 137 133 148 142 134 140 138 Distribution =B2-SBS22)/$B$25 =(B3-SBS22)/SBS25 =B4-SBS22)/$B$25 =B5-SBS22)/SBS25 =B6-SBS22)/SBS25 =B7-SBS22)/SBS25 =(B8-SBS22)/SBS25 =B9-SBS22)/SB$25 =B10-SBS22/SBS25 =B11-SB$22)/SB$25 =B12-SBS22)/SBS25 =(B13-SBS22)/SBS25 =B14-SBS22)/SBS25 =B15-SBS22)/SBS25 =B16-SB$22)/SB$25 =(B17-SB$22)/SB$25 =(B18-SBS22)/SBS25 =B19-SBS22)/SBS25 =(B20-SB$22)/SB$25 =B21-SBS22)/SBS25 10 9 11 10 1211 13 12 14 13 15 14 16 15 17 16 18 17 19 18 20 19 21 20 22 Mean 23 Median 24 Mode 146 141 138 140 138 139 142 =AVERAGEB2-B21) =MEDIANB2-B21) =MODE(B2:B21) 25 Statndard devaiation =STDEV(B2:B21) Figure 3: Formulated spreadsheet The output is shown below:

3 А 1 S. No. Control level 1 Distribution 2 1 145 0.9 135 -1.2 143 0.5 146 1.1 151 2.2 6 136 -1.0 137 -0.7 8 133 -1.6 9 148 1.5 11 10 142 0.3 12 11 134 -1.4 13 12 140 -0.1 14 13 138 -0.5 15 14 146 141 0.1 17 16 138 -0.5 18 17 140 -0.1 19 18 138 20 19 139 -0.3 21 20 142 0.3 22 Mean 140.60 23 Median 140 24 Mode 138 Statndard 25 devaiation Figure 4: Output of formulation 1.1 15 -0.5 4.82 On the basis of mean, median, mode, and distribution this control can be used. This is because, for many control level readings the value of distribution is negative, which means the standard deviation is below the mean value.

Scenario 2 Extract the data in table form as shown below: Glucose method A В. 105 99 100 99 100 101 107 93 102 104 110 91 101 99 89 111 100 102 102 96 99 100 95 105 98 96 105 94 97 103 108 92 98 10 11 13 101 14 15 16 98 104 95 100 17 18 19 20 97 103

1. Calculate the mean, median, mode, and standard deviation using MS excel spreadsheet as shown below: Formulate MS excel spreadsheet as shown below:

ДА Glucose method A В 100 7 5 102 96 105 97 108 92 3 1 100 105 4 2 101 99 5 3 107 6 4 93 99 98 8 6 104 9 7 110 10 8 91 94 119 101 12 10 99 103 13 11 89 14 12 111 15 13 100 98 16 14 101 17 15 98 18 16 96 19 17 99 195 20 18 100 100 21 19 22 20 105 103 23 Mean =AVERAGE(B3:B22) =AVERAGE(C3:C22) 24 Median =MEDIAN(B3:B22) =MEDLAN(C3:C22) 25 Mode =MODE(B3:B22) =MODE(C3:C22) Standard 26 deviation =STDEV(B3:B22) =STDEV(C3:C22) Figure 5: Formulated spreadsheet 102 102 104 195 197 aan The output is shown below:

B No. 1 3 5 91 97 12 13 8 9 10 11 12 Glucose method B 100 105 101 99 107 100 93 99 102 98 104 96 110 105 94 101 99 103 89 108 111 92 100 98 102 101 102 98 96 104 95 100 100 95 97 105 103 100.35 99.6 100.5 99 100 98 14 13 15 16 17 14 TO 15 18 16 19 99 17 18 21 19 22 20 23 Mean 24 Median 25 Mode Standard 26 deviation 4.10 5.70 Figure 6: Final output

Hence, for method A, the mean is 100.35), median is 100.5), mode is 100) and standard deviation is 15.07), and for method B the mean is 99.6, median is 99, mode is 98 and standard deviation is 4.10. As, the mean, median, and mode of method A are almost equal and of method B are also equal. Both methods show normal distribution. On the basis of normal distribution method, A seems more precise, as for method A the difference between mean median and mode is negligible. 4. The method B is more accurate with a mean of 99.6-100 and standard deviation of 4.10. This is because, according to the manufacturer listed mean is 100 mg/dL and SD is +3 mg/dL

The method B should be selected for the laboratory as it shows less deviation than method A. The chances of better quality are higher in case of method B when compared to method A.