1. Fall 2013Biostat 5110 (Biostatistics 511) Week 7 Discussion Section Lisa Brown Medical Biometry I

2. Fall 2013Biostat 5111 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( ) The central limit theorem: with large enough sample size, the distribution of sample means and sample proportions are approximately normal.

3. Fall 2013Biostat 5112 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

4. Fall 2013Biostat 5113 The normal distribution or “bell-shaped” curve has two parameters.  = the mean of X  = the standard deviation of X Normal Distribution Notation: X ~ N( ,  ) Cumulative distribution function (CDF) : P(X< c) Standard normal distribution Z ~ N(0,1) P(Z<1.65)

5. Fall 2013Biostat 5114 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

6. Fall 2013Biostat 5115 Obtaining values of standard normal CDF P(Z<c) STATA: disp normal(1.65).95052853 Or use normal probability tables (e.g. back of Baldi and Moore)

7. Fall 2013Biostat 5116 We want: P(Z<1.65)

8. Fall 2013Biostat 5117 Probability rules: complementary events = 1- P(Z>1.65)=1-P(Z<1.65)=1-.95=.05

9. Fall 2013Biostat 5118 Symmetry property of standard normal RVs = P(Z 1.65)=.05

10. Fall 2013Biostat 5119 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 -

11. Fall 2013Biostat 51110 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] Standard Normal Probabilities: more practice -0.502.0 -0.50 2.0 Why?

12. Fall 2013Biostat 51111 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. Summary: Finding probabilities: N(0,1) RVs

13. Fall 2013Biostat 51112 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-  )/  ~ N ( 0, 1 ) Converting to Standard Normal: Z scores

14. Fall 2013Biostat 51113 Probabilities for X~N( ,  ) Z = (X-  )/  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).

15. Fall 2013Biostat 51114 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. Summary: Finding probabilities: X~N(  ) RVs

16. Fall 2013Biostat 51115 Examples Suppose X~N(  =2,  =.5). What is P(1.5<X<2.75)? (STATA) disp normal(1.5)-normal(-1)=.77453754

17. Fall 2013Biostat 51116 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. Word Problems: approach

18. Fall 2013Biostat 51117 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)? Word problem: BP in older women

19. Fall 2013Biostat 51118 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) Systolic BP

20. Fall 2013Biostat 51119 Normal quantiles P(Z<1.65)=.95 The.95 quantile of a standard normal RV, z.95, is 1.65. In general, P(Z<z p )=p

21. Fall 2013Biostat 51120 Normal quantiles: example Suppose Z~N(0,1). What is the.8 quantile (or 80 th 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.

22. Fall 2013Biostat 51121 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.

23. Fall 2013Biostat 51122 Normal quantiles, continued. What about finding quantiles when X~N( ,  )? We use standarization method…in reverse. X has the same distribution as Z , where  Why? E(Z)=E(Z)  *  sd(X)=sd(Z)  What is the.8 quantile (or 80 th percentile) of X? P(Z<z.80 )=.8 P(Z  <z.80  )=P(X<z.80  =.8 Interpretation: There is an 80% chance that a randomly chosen X~N( ,  ) will fall below z.80 

24. Fall 2013Biostat 51123 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.

25. Fall 2013Biostat 51124 Approximating Binomial Distributions by Normal Distributions X~Binomial(n,p) Goal: What is the 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

26. Fall 2013Biostat 51125 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 disp binomial(20,12,.5)

27. Fall 2013Biostat 51126 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)

28. Fall 2013Biostat 51127 Sampling distribution of means Assume that X 1, X 2,...,X n are an independent, identically distributed sample of RVs from a distribution with mean  and variance   (sd  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?

29. Fall 2013Biostat 51128 Central limit theorem Assume that X 1, X 2,...,X n are an independent, identically distributed sample of RVs from a distribution with mean  and variance   (sd  Remarkably, regardless of the distribution of X i, as the sample size n gets large, Or, for large sample sizes, approximately

30. Fall 2013Biostat 51129 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

31. Fall 2013Biostat 51130 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  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:

32. Fall 2013Biostat 51131 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

33. Fall 2013Biostat 51132 Confidence intervals: interpretation For a given sample, the (for example) 95% confidence interval either contains the population mean  or it doesn’t!!! So it doesn’t make sense to to say that there is “a 95% probability that this interval contains  ” Rather, with repeated samples, a 95% confidence interval constructed with this method will contain  of the time.

34. Fall 2013Biostat 51133 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.

35. Fall 2013Biostat 51134 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?