We are going to use a simulation to solve the following probability problem: Roll a pair of dice and multiply the numbers on the uppermost faces. Find the probability that the product is greater than or equal to 18. First, solve this problem Now we are going to build a google doc to do https://chrome.google.com/webstore/detail/risk-solver/adplhaemogmadjoincpagfcmmigkinpp?hl=en-US is a link to the risk solver add on for google docs. Install it. Open a new google spreadsheet Go to Add-ons on top -> Risk Solver and hit Start In cell A1 enter die 1 [without the quotation marks J] In cell A2 enter die 2 In cell A3 enter product In cell A4 enter >= 18 In cell B3 enter =B1*B2 In cell B4 enter =if(B3>=18,1,0) Go to cell B1 Go to the far right under the Risk Solver window and hit Add Distribution Click on the distribution menu and choose IntUniform We need to change the parameters for this distribution (the integer uniform) to be 1 to 6. Do this then hit ok Move to cell B2 and repeat this entire Move to cell B4 Click on the Outputs tab on the far right under the Risk Solver window Click Add Output Click the green arrow under the Risk Solver window Once the simualation has ended you will see the following. Click on the window with the histogram under the Add Output heading Under the resulting output, the mean is the percentage of times the product was greater than or equal to 18. That s our answer! Phew! Take a screen shot of this output as that is the deliverable of this part of the problem. Also compare what your simulation returned versus what you answered in part (a) solving the problem For this problem we are going back to the Central Limit Theorem for means. In general it is difficult to actually show the CLT without using simulation. But for certain situations one can show it by hand. In this problem we will simulate by hand drawing samples of size 2 from a population. We will then examine the distribution of the resulting sample We will start with the following distribution function Is this a normal distribution (answer is yes/no). Calculate the E(X) and Var(X) If we take a sample of size n=2, what are all the possible combinations of samples with replacement we might have? List them-to help you I started the list, there will be 16 pairs listed when you are (1,1) , (2,1), (3,1) , (4,1), (1,2), (2,2), .. Calculate the means of each of the pairs you found in part (c). To help get you started the first few values are 1, 1.5, 2, 5, . Calculate the probability distribution of the sample mean. That is, create a table of the following form (the answers will go where One Direction is right now of course. Good thing too). Plot the distribution in part (e) [by hand or using the scatter command in Stata]. What does it sort of look like? How does it differ from the original distribution? Summarize in your own words what the Central Limit Theorem For this exercise, we are going to compute tail area probabilities, which we will use in earnest when we get to confidence Define the notation za to be such that P(Z < za ) = (1- a ) or equivalently alpha)x100%. Or equivalently, P(Z > za ) = a , where Z is a standard normal random variable (Z~N(0,1)). A similar notation for the t distribution can be used, tdf ,a For example z.025 = 1.96 as seen in the picture below. For the normal distribution, this tail area is found using the Stata command: display invnorm(1 a ) For the t distribution with df degrees of freedom, this area is found using the command display invttail(df, a ) Fill in the following table and comment on the results. a = .05 a = .025 a = .005 Standard Normal za 1.96 t (df=10) tdf ,a t (df=30) tdf ,a t (df=100) tdf ,a t (df=1000) tdf ,a The purpose of this exercise is to get you to hate Stata! Haha. No. Well, as much as I hate Stata now after doing this exercise. The purpose is to understand why we divide by n-1 when we calculate the sample variance (or standard deviation). Here s the overall plan. We want to create 500 data sets of size n, and compare the regular variance formula (dividing by n-1) and the formula where we just divide by n. We then want to see which one is closer to the true population variance. That is, define the following formulas n n s2 = 1 (x x )2 s2 = 1 (x x )2 i n n -1 i=1 i n i=1 Note that s2 = n 1 s2 . We will need this fact in a little bit. n n For our population we will use a handy 100 sided die. This is also called a uniform distribution on the integers 1 to 100. From Wiki (http://goo.gl/F9f7md) we can find that the mean of this distribution is m = 50.5and the variance is s 2 = 833.25. We want to generate 500 samples of n=5 dice rolls. Therefore we must roll the 100 sided die a total of 5?500 = 2500 times. We do this as follows. clear set obs 2500 generate die1 = 1 + int(100 * uniform()) If you type list you can see the 2500 values. Now we must group these 2500 numbers into groups of 5 each (this is the sample size of each group) in order to generate 500 samples. This is done as follows. generate dummy=group(500) This command will generate a variable called dummy next to the variable die1 that has the following values: 1 s from 1-5 2 s from 6-10 Type list to see the original outcomes of the 2500 rolls and the values of the dummy variable. Therefore the values of die1 that correspond to the values of dummy equal to 1 will be your first sample, the values of die1 that correspond to the values of dummy equal to 2 will be your second sample etc. Now that we have 500 data sets of size 5, we want to compute the standard deviation of each data set. This will give us 500 standard deviation values. We do this as follows. collapse (sd) die1, by(dummy) We now want to calculate the sample variances generate svar=die1*die1 We also want to compute the other variance formula (call it nvar). Using the relationship above generate nvar=(4/5)*svar We are now ready to compare the regular and n formulas. We will do so by comparing the mean of the 500 values of each. summarize svar nvar Variable | Obs Mean Std. Dev. Min Max -+ svar | 500 835.371 420.5599 40.3 2237.2 nvar | 500 668.2968 336.4479 32.24 1789.76 It looks like on average, the sample variance is closer to the population variance (833.25) than the formula that uses n . What happens if we increase sample size? That s for you to find out. Fill in the following table doing the above commands for n=10, 25, 50 and 100. Because you re nice, I ve put the commands into a text file called compare_sd.txt which is on the web site. You ll just have to modify the number of observations (I ve marked where in the code you need to do this), then cut and paste it into Stata. When you re done, make a table like the one below and comment on your findings. Remember that what we want to study is the average of the 500 standard deviations computed each of two ways. Results of Simulation with 500 trials for each sample size n=5 n=10 n=25 n=50 n=100 s 2 833.25 833.25 833.25 833.25 833.25 s2 833.25 2 sn 668.2968 Now that you re experienced with Stata, we re going to do another simulation (goodie!) to study the Central Limit The Stata code in file simul_clt.txt generates 500 samples, where each sample consists of rolling a 6 sided die 80 times. The variable xbar is the average of the 80 rolls of the die each time, for 500 separate times. Cut and paste this code into Stata. Answer the following questions: The population from where the samples were taken has =3.5 and a-=1.71. According to the central limit theorem what is the distribution of the sample mean (n=80). Construct the histogram of these 500 sample means and ask Stata to draw a normal curve on top of that. Here is the command: histogram xbar, normal . What do you observe? Compute the mean of these 500 means. What do you find? Compute the standard deviation of these 500 sample What do you find? Find the probability that the sample mean x is larger than 3.6. Remember has =3.5 and a-=1.71, n=80. What proportion of these 500 sample means is larger than 3.6 (count how many sample means are larger than 3.6 out of the 500)? It will be helpful if you sort the sample means by using . sort xbar . Compare this empirical probability with the probability that you found in the previous The reading speed of second grade students is approximately normal, with a mean of 88 words per minute (wpm) and a standard deviation of 12 What is the probability a randomly selected student will read more than 95 words per minute? What is the probability that a random sample of 12 second grade students results in a mean reading rate of more than 95 words per minute? What is the probability that a random sample of 24 second grade students results in a mean reading rate of more than 95 words per minute? What effect does increasing the sample size have on the probability? Provide an explanation for this A teacher instituted a new reading program at school. After 10 weeks in the program, it was found that the mean reading speed of a random sample of 20 second grade students was 92.8 wpm. What might you conclude based on this result? There is a 5% chance that the mean reading speed of a random sample of 20 second grade students will exceed what value? The shape of the distribution of the time required to get an oil change at a 10-minute oil-change facility is unknown. However, records indicate that the mean time for an oil change is 11.6 minutes, and the standard deviation for oil-change time is 3.1 minutes. To compute probabilities regarding the sample mean using the Central Limit Theorem, what size sample would be required? What is the probability that a random sample of n = 40 oil changes results in a sample mean time of less than 10 minutes? Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 40 oil changes between 10 a.m. and 12 p.m. Treating this as a random sample, what mean oil-change time would there be a 10% chance of being at or below? This will be the goal established by the The mean selling price of senior condominiums in Green Valley over a year was $215,000. The population standard deviation was $ 25,000. A random sample of 100 new unit sales was obtained. What is the probability that the sample mean selling price was more than $210,000? What is the probability that the sample mean selling price was between $ 213,000 and $ 217,000? Suppose that, after you had done these calculations, a friend asserted that the population distribution of selling prices of senior condominiums in Green Valley was almost certainly not normal. How would you respond? The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer randomly selected 36 swans and loaded them into his truck. What is the probability that this flock of swans weighs > 1000 pounds? Let X1, X2, X3, and X4 be a random sample of observations from a population with mean and variance ??2. Consider the following estimator of : 1 = 0.35 X1 + 0.25 X2 + 0.10 X3 + 0.30 X4. What is the expected value of 1? What is the variance of 1? How does the variance compare to X , the usual estimate of m ? (that is, which variance is larger and hence which estimator would you prefer?) An insurance company is interested in the average claim on its auto insurance policies. Using 36 randomly selected claims, it finds the mean claim to be $1,270 with a standard deviation of $421. Construct a 95 percent confidence interval for the mean claim on all A random sample of the luggage of 49 passengers of Jet Blue finds that the mean weight of the luggage is 47 pounds with a standard deviation of 8 pounds. Construct a 95 percent confidence interval for the mean weight of Jet Blue Airlines For this problem we are going to use class survey data from a previous offering of Stat Load the data into Stata using the command: use http://people.fas.harvard.edu/~mparzen/stat104/stat111_survy After the data is loaded, enter the Stata command sort female Find a 95% confidence interval for the weight Now enter the command ci weight,by(female). How do these intervals compare with each other and the interval found in (a)? Find a 95% confidence interval for the sleep variable; how many hours of sleep students usually get per Are you inside the interval? The variable looks is what percentage of people at Harvard are better looking than you. Find a 95% confidence interval for this variable for men and women separately. Do the intervals appear that different? The variable haircut is what do you usually pay for a haircut. Find a 95% confidence interval for this variable for men and women separately. Do the intervals appear that different? Find a 95% confidence interval for the variable texts, the number of texts you send per Are you inside this interval? Strictly speaking, what is the best interpretation of a 95% confidence interval for the mean? If repeated samples were taken and the 95% confidence interval was computed for each sample, 95% of the intervals would contain the population A 95% confidence interval has a 95 probability of containing the population mean. 95% of the population distribution is contained in the confidence
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