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Medical Biometry I

Medical Biometry I. ( Biostatistics 511) Week 7 Discussion Section Lisa Brown. T he Normal Distribution. Many “Real world” measurements, such as IQ and height can be modeled was normal random variables (RVs).

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Medical Biometry I

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  1. Medical Biometry I (Biostatistics 511) Week 7 Discussion Section Lisa Brown Biostat 511

  2. The Normal Distribution • Many “Real world” measurements, such as IQ and height can be modeled was normal random variables (RVs). • Some RVs can have distributions that are approximately normal (provided certain conditions apply) • Binomial(n, p) • Poisson(l) • The central limit theorem: with large enough sample size, the distribution of sample means and sample proportions are approximately normal. Biostat 511

  3. Skills and Concepts • The standard normal distribution and z-scores • Finding probabilities • Finding quantiles, given probabilties • Word problems • Using the normal approximation of the binomial distribution. • Distribution of sample means and the central limit theorem • Forming Confidence intervals for population means Biostat 511

  4. Normal Distribution • The normal distribution or “bell-shaped” curve has two parameters. •  = the mean of X • = the standard deviation of X • Notation: X ~ N(, ) • Cumulative distribution function (CDF) : P(X< c) • Standard normal distribution Z ~ N(0,1) P(Z<1.65) Biostat 511

  5. Normal PDF and CDF Total area of a PDF=1 So P(Z<1.65)=.95 is the area of the shaded region. Interpretation: 95/100 samples of Z will be less than or equal to 1.65 Biostat 511

  6. Obtaining values of standard normal CDF P(Z<c) STATA: dispnormal(1.65) .95052853 Or use normal probability tables (e.g. back of Baldi and Moore) Biostat 511

  7. We want: P(Z<1.65) Biostat 511

  8. Probability rules: complementary events = 1- P(Z>1.65)=1-P(Z<1.65)=1-.95=.05 Biostat 511

  9. Symmetry property of standard normal RVs = P(Z<-1.65)=P(Z>1.65)=.05 Biostat 511

  10. Probabilities of intervals = - P(-1.65<Z<1.65)=P(Z<1.65) - P(Z<-1.65) = .95-.05=.90 For a standard normal RV, 90% of values fall between -1.65 and 1.65 Biostat 511

  11. Standard Normal Probabilities: more practice P[Z < 1.65] = 0.9505 P[Z > 0.5] = 1-P[Z < 0.5] = 0.3085 P[-1.96 < Z < 1.96] = P[Z < 1.96] - P[Z < -1.96] = .95 P[-0.50 < Z < 2.0] = P[Z <2.0] - P[Z <-0.50] 2.0 -0.50 Why? -0.50 2.0 Biostat 511

  12. Summary: Finding probabilities: N(0,1) RVs Step 1. Draw picture of area corresponding to probability. Step 2. Use probability rules and tables or STATA to find quantities in (1). Step 3. Get the answer. Biostat 511

  13. Converting to Standard Normal: Z scores Q: This solves the problem for the N(0,1) case. How do we do calculate normal probabilities when the mean is not 0 and the standard deviation is not equal to 1? A:Any normal random variable can be transformed to N(0,1) E(X- ) = 0 V(X- ) = V(X) = 2 V( (X- )/ ) =(1/2)*V(X)=1 Linear transformations of normal random variables are still normal. So Z = (X-m)/s ~ N ( 0 , 1 ) Biostat 511

  14. Probabilities for X~N(,) Z = (X-m)/s is a rescaled and shifted version of X—like going from Fahrenheit to Celsius. In other words, the probability that X<2.822 is the same as the probability Z<1.65, since (X-2)/.5 ~N(0,1). Biostat 511

  15. Summary: Finding probabilities: X~N(m,s) RVs Step 0. Draw picture of area corresponding to probability. Step 1. Re-express probability statement about X as statement about Z by standardizing. Step 2. Use probability rules and tables or STATA to find quantities in (1). Step 3. Get the answer. Biostat 511

  16. Examples Suppose X~N(m=2,s=.5). What is P(1.5<X<2.75)? (STATA) disp normal(1.5)-normal(-1)=.77453754 Biostat 511

  17. Word Problems: approach • Define the random variable in words. • Is it normally distributed? What is the mean and standard deviation? • What is the event and corresponding probability statement? • Draw picture of area corresponding to probability. • Re-express probability statement about X as statement about Z by standardizing. • Use probability rules and tables or STATA to find probabilities. • Get the answer. Biostat 511

  18. Word problem: BP in older women Suppose a clinically accepted value for mean systolic blood pressure in females, aged 65-74 is 133 mmHg and the standard deviation is 20 mmHg. If a 70-year-old- woman is selected at random from the population, what is the probability that her systolic blood pressure is equal to or less than 120 mmHg? X = systolic BP in woman age 65-74.  = 133  = 20 What is P(X< 133)? Biostat 511

  19. Systolic BP Example Suppose a clinically accepted value for mean systolic blood pressure in females, aged 65-74 is 133 mmHg and the standard deviation is 20 mmHg. If a 70-year-old- woman is selected at random from the population, what is the probability that her systolic blood pressure is equal to or less than 120 mmHg? STATA: display normal(-0.65) Biostat 511

  20. Normal quantiles P(Z<1.65)=.95 The .95 quantile of a standard normal RV, z.95, is 1.65. In general, P(Z<zp)=p Biostat 511

  21. Normal quantiles: example Suppose Z~N(0,1). What is the .8 quantile (or 80th percentile) of Z? P(Z<z.80)=.8 STATA: display invnorm(.8) .84162123 Interpretation: There is an 80% chance that a randomly chosen Z~N(0,1) will fall below .84. Biostat 511

  22. Normal quantiles: tables P(Z<z.80)=.8…Find values of z with p closest to .8 From the table, P(Z<.84)=.7995 and P(Z<.85)=.8023 So the .8th quantile is approximately .845. Biostat 511

  23. Normal quantiles, continued. What about finding quantiles when X~N(m,s)? We use standarization method…in reverse. X has the same distribution as Zs+m, where Z~N(0,1) Why? E(Z)=E(Z)s+m=0*s+m=m sd(X)=sd(Z)s=s What is the .8 quantile (or 80th percentile) of X? P(Z<z.80)=.8 P(Zs+m<z.80s+m)=P(X<z.80s+m)=.8 Interpretation: There is an 80% chance that a randomly chosen X~N(m,s) will fall below z.80s+m=.84*s+m. Biostat 511

  24. Normal quantiles: example Suppose a clinically accepted value for mean systolic blood pressure in females, aged 65-74 is 133 mmHg and the standard deviation is 20 mmHg. Between what two blood pressure readings will 80% of all systolic blood pressures for 65-74-year-old women lie? We want the .1 and .9 quantiles of X, since 80% of all values lie in this range. P(z.1<Z<z.9)=.80 P(20z.1+133<20Z+133<20z.9+133)=.8 P(20z.1+133<X<20z.9+133)=.8 P(20*(-1.2816)+133<X<20*1.2816+133) So 80% of BP readings will fall between 107.4 and 158.6. Biostat 511

  25. Approximating Binomial Distributions by Normal Distributions X~Binomial(n,p) Goal: What is the P(X<c) or P(X>c)? Tail probabilities using the binomial distribution can be tedious to compute, especially by hand! If np and n(1-p) are large enough (>10), then approximately Biostat 511

  26. Example If np and n(1-p) are large enough (>10), then approximately X~Binomial(n=200, p=.4). What is P(X<70)? 200*.4>10 and 200*.6>10, so, approximately Exact calculation P(X<70)=.0843 STATA dispbinomial(20,12,.5) Biostat 511

  27. Example What happens if np and n(1-p) are not large enough? The normal approximation can be terrible! X~Binomial(n=10, p=.1). What is P(X<1)? Does not meet “rule of thumb” for normal approx: np=1, n(1-p)=9. If we assume it anyway, Exact calculation P(X<1)= .74 STATA: display binomial(10,1,.1) Biostat 511

  28. Sampling distribution of means Assume that X1, X2,...,Xn are an independent, identically distributed sample of RVs from a distribution with mean m and variance s2 (sds) . The sample mean is another random variable So as n gets, bigger, the standard deviation of the sample mean goes down. If sd(X) =10, what is the sd of the the sample mean when n=100? Biostat 511

  29. Central limit theorem Assume that X1, X2,...,Xn are an independent, identically distributed sample of RVs from a distribution with mean m and variance s2 (sds) . Remarkably, regardless of the distribution of Xi, as the sample size n gets large, Or, for large sample sizes, approximately Biostat 511

  30. Central limit theorem at work The CLT is very powerful: no matter how skewed the distribution of X, the distribution of a sample mean will approach normality with increasing n. How large does n need to be for the normal approximation to be good? It depends on the distribution of X. Distribution of sample mean for different N Biostat 511

  31. Confidence intervals One goal of statistical inference is to estimate population means. We use the sample mean, as a point estimate. This estimate is better for larger n, since is less variable and closer to m with increasing n. Confidence intervals allow us to express the uncertainty about our estimate of the mean, by citing a range of values rather than a single point. We construct a “p-percent” confidence interval for mu as follows: Biostat 511

  32. Finding the critical value for a “p %” confidence interval -We need to find the standard normal quantile, z*, such that the shaded area P(|Z|<z*)=p. -This corresponds to the 1-(1-p)/2 quantile (see picture)! -For a 90% confidence interval, 1-(1-p)/2=1-.10/2=.95, so z* is z.95.=1.645 What about 95% confidence? z*=z.975=1.96. That is, each of the tail regions have area (1-p)/2. So z* corresponds to the 1-(1-p)/2 quantile of the standard normal Distribution. p Right tail probability (1-p)/2 Biostat 511

  33. Confidence intervals: interpretation For a given sample, the (for example) 95% confidence interval either contains the population mean m or it doesn’t!!! So it doesn’t make sense to to say that there is “a 95% probability that this interval contains m.” Rather, with repeated samples, a 95% confidence interval constructed with this method will contain m 95% of the time. Biostat 511

  34. Confidence interval Example Your goal is to estimate the mean of systolic BP in a population of women 65-75. You collect a sample of 100 women. Suppose you know that the standard deviation for systolic BP in the population is 20. The mean BP in your sample is 125. Construct and interpret a 95% confidence interval for the population mean BP. For a 95% CI, the critical value z*=1.96 95% Confidence interval: [125-1.96*20/10, 125+1.96*20/10]= [121.08, 128.92]. Interpretation: with repeated samples, 95% of intervals formed with this method would contain the true mean BP. Biostat 511

  35. Confidence interval: discussion What affects the width of the confidence interval? Confidence intervals depend on the CLT and normal approximation for the sample mean’s distribution. For small n, is this still a good approach? Biostat 511

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