Zhijin Wu PhD

Formula for midtrem 2 (44.5 KB)
(IT has some extra formulas you won't need yet)
Lecture 1 (73.1 KB)
Introduction to statistics
Lecture 2 (0.1 MB)
exploratory analysis and summary statistics. (SMMR Chapter 2)
lecture 3 (0.2 MB)
Introduction to probability (SMMR 3.1-3.3//MSDA Chap. 1). A review of permutation/combination may be helpful before the class http://www.youtube.com/watch?v=buyevw1flZk
Lecture 4 (0.2 MB)
discrete random variables (SMMR 3.4,3.6,3.7//MSDA 2.1)
Lecture 5 (0.6 MB)
Oct 4 (SMMR 3.8 //MSDA2.2.1, 2.2.3)
Lecture 6 (0.1 MB)
mean and variance of a random variable (MSDA 4.1-4.3)
Lecture 6b (0.1 MB)
lecture 7 (85.5 KB)
likelihood （MSSD 9.1,9.2)
Lecture 8 (2.7 MB)
Law of large numbers and Central Limit Theorem (MSDA Chapter 5 & 7)
Lecture 9 (0.3 MB)
Confidence Intervals (SMMR Chapter 4; MSDA Chapter 7)
Lecture10 (0.2 MB)
SMMR Chapter 4
lecture13a (0.1 MB)
extra lecture on ANOVA
lecture13b (0.1 MB)
correlation and regression

hw1.pdf (17.8 KB)
Due 9/18.
hw2 (23.3 KB)
Due Oct 2. (Before 5pm, if you cannot turn in at lecture hour)
hw3 (22.6 KB)
Due Oct 11
hw4 (33.6 KB)
Due Oct 25.
hw6 (0.2 MB)
Due 11-29
Lab3 (0.1 MB)
Probability Exercises

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 More papers you can see examples You should have access to these papers as long as you are on Brown Network. http://www.nejm.org/doi/full/10.1056/nejmoa0909859 See Table 1 for comparison of baseline characteristics. {12/6/2012 1:15 PM} {0 Comments}
 More Confidence Interval Examples Below are some additional examples on confidence interval. You can use these as exercises. To check your answer, highlight the blank space after each question to view the solution. Q:  Suppose a student measuring the boiling temperature of a certain liquid observes the readings (in degrees Celsius) 102.5, 101.7, 103.1, 100.9, 100.5, and 102.2 on 6 different samples of the liquid. He calculates the sample mean to be 101.82. If he knows that the measurements follow a normal distribution and the standard deviation for this procedure is 1.2 degrees, what is the confidence interval for the population mean at a 95% and 90% confidence level? A: Notice that "he knows the standard deviation". The critical value for a 95% confidence interval is 1.96, where (1-0.95)/2 = 0.025. A 95% confidence interval for the unknown mean is ((101.82 - (1.96*0.49)), (101.82 + (1.96*0.49))) = (101.82 - 0.96, 101.82 + 0.96) = (100.86, 102.78). For confidence level 90%,  (1-90%)/2 = 0.05. The critical value z* for this level is equal to 1.645, so the 90% confidence interval is ((101.82 - (1.645*0.49)), (101.82 + (1.645*0.49))) = (101.82 - 0.81, 101.82 + 0.81) = (101.01, 102.63) Q: A dataset on "Normal Body Temperature, Gender, and Heart Rate" contains 130 observations of body temperature, along with the gender of each individual and his or her heart rate. A summary of descriptive statistics is provided below: `Descriptive Statistics Variable N Mean Median Tr Mean StDev SE Mean TEMP 130 98.249 98.300 98.253 0.733 0.064 Variable Min Max Q1 Q3 TEMP 96.300 100.800 97.800 98.700 ` Find a 95% confidence interval for the mean body temperature. A:   To find a 95% confidence interval for the mean based on the sample mean 98.249 and sample standard deviation 0.733, first find the 0.025 critical value t* for 129 degrees of freedom. This value is approximately 1.962, the critical value for 100 degrees of freedom (found in Table E in Moore and McCabe). The estimated standard deviation for the sample mean is 0.733/sqrt(130) = 0.064, the value provided in the SE MEAN column of the MINITAB descriptive statistics. A 95% confidence interval, then, is approximately ((98.249 - 1.962*0.064), (98.249 + 1.962*0.064)) = (98.249 - 0.126, 98.249+ 0.126) = (98.123, 98.375).     Q:The following data (n = 10) were drawn from a normal population -4.26549, -4.50909, 1.26475, 1.42241, 2.73875, 11.954, 3.61592, -9.68883, -2.96558, -3.48133 The sample mean is  μ = −0.39145 and the sample variance is  35.431. Construct a 95% confidence interval. A:  We begin with normal distribution, so we don't need CLT to know that the sample mean also follows normal distribution. But we do not know the variance and have to estimate it from the data. Thus we use t-distribution with degree of freedom n-1=9. t (0.025,df=9)= 2.2622 CI: (−4.65, 3.87) Q: The same experiment was repeated but this time for n = 100.   Now the sample mean is μ= 2.1373 and the sample variance is 31.855. What is the 95% confidence interval now? A: when sample size changes your degree of freedom changes. t (0.025,df=99) = 1.9842 CI: [1.02, 3.26] Q: A candidate in a two-person election commissions a poll to determine who is ahead. The pollster randomly chooses 500 registered voters and determines that 260 out of the 500 favor the candidate. Estimate the support rate for this candidate in all registered votes and construct a 95% confidence interval for this rate. A:  Use the formula for estimating proportions. p.hat=0.52. n=500 Lower limit: 0.52 - (1.96)(0.0223) - 0.001 = 0.475 Upper limit: 0.52 + (1.96)(0.0223) + 0.001 = 0.565 CI: (.475,.565) NOTE: The solution was taken from a book which added a little continuity correction .5/N to expand the CI a little bit, because the normal distribution is an approximation. This is not necessary for your exercise/hw/exam.   0.52 +/- (1.96)(0.0223) is a perfectly acceptable. Q: {11/2/2012 10:02 AM} {0 Comments}