diagnosing spina bifida诊断脊柱裂的问题课件

C6, L1, S1 Problem: Diagnosing Spina Bifida The procedure of amniocentesis involves drawing a sample of the amniotic fluid that surrounds an unborn child in its mothers womb. High concentration of alpha fetoprotein can indicate the condition spina bifida. Concentration of alpha fetoprotein tends to increase with the size of the foetus. Amniocentesis results in miscarriage for 1%. Preliminary tests involve measuring the level of alpha fetoprotein in the mothers urine. C6, L1, S2 Problem: Diagnosing Spina Bifida For mothers with normal foetuses, the mean level of alpha fetoprotein is 15.73 moles/litre with a standard deviation of 0.72 moles/litre. For mothers carrying foetuses with spina bifida, the mean is 23.05 and the standard deviation is 4.08. In both groups the distribution of alpha fetoprotein appears to be approximately Normally distributed. 23.05 15.73 C6, L1, S3 Problem: Diagnosing Spina Bifida To operate a diagnostic test for spina bifida, set a threshold concentration of alpha fetoprotein, T, say. If the alpha fetoprotein level is below T, then the foetus is diagnosed as not having spina bifida. If the level is above T, then further testing is required. C6, L1, S4 Problem: Diagnosing Spina Bifida If T was set at 17.80 moles/litre: What is the probability that a foetus with spina bifida is correctly diagnosed? What is the probability that a foetus not suffering from spina bifida is correctly diagnosed? If they wanted to ensure that 99% of foetuses with spina bifida were correctly diagnosed, at what level should they set T ? What are the implications of setting T at this level? C6, L1, S5 Chapter 6 Continuous Random Variables If a random variable, X, can take any value in some interval of the real line it is called a continuous random variable. Eg Hg levels, height, weight, alpha fetoprotein concentration, cell radius, etc. (i.e. usually measures ) C6, L1, S6 6.1 pages 231-233 The Standardized Histogram Example: Dietary Carbohydrate in the Workforce The average daily intake of carbohydrate in the diet of 5929 people. C6, L1, S7 200400600800 Carbohydrate (g/day) 0 .002 .000 .004 The Standardized Histogram The histogram of the data shows the carbohydrate intake: Is unimodal (modal class 200 225 g/day) C6, L1, S8 200400600800 Carbohydrate (g/day) 0 .002 .000 .004 The Standardized Histogram The histogram of the data shows the carbohydrate intake: Skewed to larger values (skewed right) C6, L1, S9 200400600800 Carbohydrate (g/day) 0 .002 .000 .004 The Standardized Histogram The histogram of the data shows the carbohydrate intake: Has huge variability (highest consumers more than 10 times that of lowest consumers) C6, L1, S10 Area between a = 225 and b = 375 shaded 0600800 225 .002 .000 .004 375 Shaded area = 0.483 (Corresponds to 48.3% of observations) The Standardized Histogram C6, L1, S11 The standardized histogram adjusts the height of the rectangle or bar to relative freq. or proportion divided by width so that Area = Estimated Probability. 0600800 225 .002 .000 .004 375 Shaded area = 0.483 (Corresponds to 48.3% of observations) The Standardized Histogram C6, L1, S12 i.e. The area of the ith rectangle tells us what proportion of the data lie in the ith class interval. 0600800 225 .002 .000 .004 375 Shaded area = 0.483 (Corresponds to 48.3% of observations) The Standardized Histogram C6, L1, S13 For a standardized histogram: The vertical scale is : Relative frequency / interval width (density scale) Total area under the histogram = 1 The proportion of the data between a and b is the area under the histogram between a and b. The Standardized Histogram C6, L1, S14 With approximating curve Carbohydrate (g/day) 2 0040 06 008 00 .004 .002 0 The Standardized Histogram C6, L1, S15 Area between a = 225 and b = 375 shaded Shaded area = .486 (cf. area = .483 for histogram) .04 0600800 225 375 .002 This area is calculated to be 0.486 and is very close to the proportion of people who had carbohydrate intake of between 225 and 375 g/day. The Standardized Histogram C6, L1, S16 Radius of Maliginant Tumor Cells In JMP select Histogram Options Density Axis to create a standardized histogram The histogram on the left is for cell radii of malignant tumor fine needle aspirations in the breast cancer study from your 2nd assignment. X = radius of a randomly selected malignant tumor cell We estimate that, P(14 Density Axis to create a standardized histogram The histogram on the left is AFP levels found in the urine of mothers carrying a fetus with spina bifida. X = AFP level of random select mother carrying fetus with spina bifida. We estimate that, P(22.5 17.8) or P(19 -1.29) = 1 - P(Z 17.8) = P(Z z-score for Y = 17.8) Original problem: Diagnosing Spina Bifida 23.05 15.73 z-score = (17.8 23.05)/4.08 = -5.25/4.08 = -1.29 -1.290 C6, L1, S39 Original problem: Diagnosing Spina Bifida If they wanted to ensure that 99% of foetuses with spina bifida were correctly diagnosed, at what level should they set T ? 23.05 15.73 Find a value T so that if Y Normal (23.05, 4.08) we will have P(Y T) = .9900 or P(Y T) = .0100 First find the z-score associated with T by finding z so that P(Z z) = .0100 From Normal Table we find P(Z -2.33) = .0100 thus T = + s x z = 23.05 4.08 x 2.33 = 13.54 T = 13.54 ensures 99% of foetuses with spina bifida will be identified. This probability is called the sensitivity C6, L1, S40 Standard Normal Probabilities in JMP Normal Probability Calculator.JMP from Tutorials section of course website. Here it is ready to calculate probabilities for the standard normal d