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Overview
Unit 8: Overview and Integration Topics: Descriptive statistics Probability Sources of variability Describing and summarizing variability The Central Limit Theorem for averages and for sums. Learning Objectives: Have a better understanding of the relation between descriptive statistics, probability, and inferential statistics. Distinguish between the different uses of the concept of variability. Integrate the tools that were given in the first part of the book in order to solve complex problems. Check List: Read Chapter 8 of the textbook. Actively follow the examples of R code presented in Chapter 8. Review the material of Units 1-7. Complete the Practice Quiz and the Graded Quiz.
Complete the Assignment for Unit 8. Post your answer to the question in your Discussion Forum and follow it up with approximately 3-4 comments to other students posts. Rate other students posts. Post your activities throughout the week in the Learning Journal and complete the Learning Journal vue d'ensemble Unité 8: Vue d'ensemble et de l'intégration sujets: Les statistiques descriptives probabilité Sources de variabilité Décrire et résumer la variabilité Le théorème central limite pour les moyennes et pour des sommes. Objectifs d'apprentissage: Avoir une meilleure compréhension de la relation entre les statistiques descriptives, la probabilité et statistiques déductives. Distinguer les différents usages de la notion de variabilité. Intégrer les outils qui ont été donnés dans la première partie de l'ouvrage afin de résoudre des problèmes complexes. Liste de vérification: Lisez le chapitre 8 du manuel. Suivre activement les exemples de code R présentés au chapitre 8. Revoyez le matériel de parts de 1-7. Remplissez le Quiz pratique et l'Graded Quiz. Remplissez la cession de l'Unité 8. Diffusez votre réponse à la question dans votre forum et le suivi avec environ 3-4 commentaires à d'autres étudiants messages. Taux autres étudiants messages. Publiez vos activités tout au long de la semaine dans le Journal d'apprentissage et de compléter le Journal d'apprentissage.
Estimating the Expectation A measurement follows the Normal distribution with standard deviation that is equal to 15 and an unknown expectation μ. Two statisticians propose two distinct ways to estimate the unknown quantity μ with the aid of a sample of size 36. Statistician A proposes to use the sample average as an estimate. Statistician B proposes to use the sample median instead. In order to choose between the two options they agree to prefer the statistics that has a smaller variance (with respect to the sampling distribution). Tasks 1-9 refer to this problem of comparing the two statistics to each other. Normal Approximation of the Sampling Distribution of a Sum Suppose that the expected number of phone calls that are handel by a switchboard in each second is 5.35. Assume that the distribution of the number of phone calls per second follows the Poisson distribution. Tasks 10-12 refer to this information. Submitting the Assignment For the assignment you should complete 12 tasks. Tasks 1-9 refer to the information associated with the estimating the expectation and Tasks 10-12 refer to the application of the Normal approximation of the sampling distribution of a sum. This assignment has the format of a quiz and it is computer graded. The tasks are described in the Form titled "Assignment Unit 8". The answer to each task involves the selection of the right choice (in a multiple choice question) or the entering of a numerical value at the designated location. Notice: Unlike the Quizzes, you will not be given a chance to correct a wrong answer after submitting the assignment. Hence, we recommend that you check your answers carefully before submitting the assignment. affectation Estimation de la Attente Une mesure suit la distribution normale avec un écart type égal à 15 et une attente inconnu μ. Deux statisticiens proposent deux façons distinctes pour estimer la quantité inconnue μ à l'aide d'un échantillon de taille 36. Un statisticien propose d'utiliser la moyenne de l'échantillon comme une estimation. Statisticien B propose d'utiliser la médiane de l'échantillon à la place. Afin de choisir entre les deux options qu'ils conviennent à préférer les statistiques qui a une variance plus faible (par rapport à la distribution d'échantillonnage). Tâches 1-9 font référence à ce problème de la comparaison des deux statistiques à l'autre.
Approximation normale de la distribution d'échantillonnage d'une somme Supposons que le nombre prévu d'appels téléphoniques qui sont Handel par un tableau dans chaque seconde est 5,35. On suppose que la distribution du nombre d'appels par seconde suit la loi de Poisson. Tâches 10-12 se réfèrent à cette information. Envoi de la cession Pour l'affectation vous devez remplir 12 tâches. Tâches 1-9 font référence à l'information associée à l'estimation de l'espérance de tâches et de 10 à 12 se rapportent à l'application de l'approximation normale de la distribution d'échantillonnage d'une somme. Cette affectation a le format d'un quiz et il est classé ordinateur. Les tâches sont décrits dans le formulaire intitulé «Affectation Unité 8". La réponse à chaque tâche implique la sélection du bon choix (dans une question à choix multiples) ou l'entrée d'une valeur numérique à l'endroit désigné. AVIS: Contrairement aux Quiz, vous ne serez pas donné une chance de corriger une mauvaise réponse après le dépôt de la cession. Par conséquent, nous vous recommandons de vérifier vos réponses avant de soumettre la cession.
Discussion Forum Question
In this course we have learned many subjects. Most of these subjects, especially for those that had no previous exposure to statistics, where unfamiliar. In this forum we would like to ask you to share with us the difficulties that you encountered. What was the topic that was most difficult for you to grasp? In your opinion, what was the source of the difficulty? When forming your answer to this question we will appreciate if you could elaborate and give details of what the problem was. Pointing to deficiencies in the learning material and confusing explanations will help us improve the presentation for the future application of this course.
Question 2: What would you expect to happen to the histogram for all.obs if we collected 200 more samples (from the same store) that each contain 15 observations? If you want to test your theory, you could create additional samples using an R command something like this: x <- rexp(15, rate=3), or you could build a simulation similar to one in the book (Yakir, 2011 pp. 117-118). If you follow the simulation in the book, remember to change the sample size to 15 instead of 10, then ignore the samples sizes for 100 and 1000. x01 <- c(.01, .01, .04, .31, .64, .21, .51, .52, .05, .12, .22, .01, .05, .69, .09); x02 <- c(.15, .49, .33, .12, .50, .07, .81, .28, .21, .10, .31, .75, .09, .51, .19); x03 <- c(.26, .19, .18, .51, .95, .08, .03, .28, .30, .48, .02, .31, .06, .93, .10); x04 <- c(.15, .56, .67, .39, .03, .54, .27, .03, .63, .21, .68, .58, .05, .35, .31); x05 <- c(.68, .30, .46, .03, .18, .04, .18, .33, .20, .05, .16, .34, .45, .06, .06); x06 <- c(.07, .11, .53, .45, .66, .63, .15, .06, .12, .01, .01, .14, .84, .49, .57); x07 <- c(.34, .46, .43, .96, .49, .21, .05, .14, .12, .09, .42,1.14, .08, .06, .76); x08 <- c(.10, .35, .40, .87, .39, .19, .45, .16, .22, .33, .26, .54, .22, .46, .53); x09 <- c(.52, .04, .41, .20, .05, .03, .01, .03, .22, 1.41, .38, .81, .12, .04, .33); x10 <- c(.82, .01, .11, .30, .84, .04, .08, .05, .12, .14, .04, .35, .00, .17, .07); x11 <- c(0.08, .40, .41, .61, .07, .33, .02, .08, .12, .31, .21, .41, .84, .59, .12); x12 <- c(0.03, .12, .03, .02, .31, .14, .84, .20, .02, .01, .45, .62, .26, .40, .18); x13 <- c(0.59, .09, .05, .13, .32, .03, .57, .84, .00, .27, .11, .43, .70, 1.20, .31); x14 <- c(0.49, .20, .12, .69, .15, .77, 1.33, .08, .28, .78, .25, .23, .10, .40, .02); x15 <- c(0.26, .00, .31, .12, .01, .82, .40, .06, .21, .08, .04, .61, .03, .21, .96); x16 <- c(0.02, .41, .01, .03, .28, .31, .52, .70, .33, 1.47, .03, .36, .21, .77, .29); x17 <- c(0.01, .12, .00, .18, .22, 1.22, .98, .43, .25, .23, .34, .26, .02, .20, .02); x18 <- c(0.40, .12, .41, .29, .21, .80, .05, .09, .26, .01, .23, .45, 1.23, .23, .10); x19 <- c(0.03, .56, .03, .41, .33, .81, .00, .03, .44, .52, .62, .44, .60, 1.11, .14); x20 <- c(0.00, .04, .04, .06, .11, .44, .44, .73, .03, .37, .04, .16, .48, .32, .08);
mean(x01); mean(x02); mean(x03); mean(x04); mean(x05); mean(x06); mean(x07); mean(x08); mean(x09); mean(x10); mean(x11); mean(x12); mean(x13); mean(x14); mean(x15); mean(x16); mean(x17); mean(x18); mean(x19); mean(x20); samp.dist <- c(0.232 , .3273333 , .312 , .3633333 , .2346667 , .3226667 , .3833333 , .3646667 , .3066667 , .2093333, .2226667 , .242 , .2746667 ,. 2986667 , .3066667 , .3253333 , .376 , .3826667 , .3926667 , .4046667) all.obs <- c(x01, x02, x03, x04, x05, x06, x07, x08, x09, x10, x11, x12, x13, x14, x15, x16, x17, x18, x19, x20) hist(all.obs) hist(samp.dist)
> exp. hist(exp.15) When I typed hist(exp.15), I got a more distributed data than the “samp.dist” histogram. Then, instead of 200 samples, I did 20000 samples. Just like this: > exp.15 <- rep(0,20000) > for(i in 1:20000)
- {
- X.samp.15 <- rexp(15,.5)
- exp.15[i] <- mean(X.samp.15)
- } > hist(exp.15) Here, when I typed hist(exp.15), I got an even more distributed data. The fact I computed the means of each sample and make a data of just means helped me to get a more normal distribution.
I also tried to find out what would happen if I do the same process to answer question 2. So I entered the following functions. Note that this time, I did not compute the mean of each sequence of numbers. exp.15 <- rep(0,200) > for(i in 1:200)
- {
- X.samp.15 <- rexp(15,.5)
- exp.15[i] <- (X.samp.15)
- } > hist(exp.15) Here, the figure seems to be chopped off on the right side. It looks similar to the “all.obs” distribution of this week's discussion. The bars on the left where the highest. I decided to repeat the same process, but this time with a million sequences of 15 numbers. > exp.15 <- rep(0,1000000) > for(i in 1:1000000)
363 words Rate: Show parent | Reply Re: MATH1280 Unit 8 Discussion by Curtis Szmania - Sunday, 4 January 2015, 5:03 PM As my suspicions suggested the all.obs histogram was similar in distribution, with the highest numbers on the left. I estimated that the bars would be higher because the sample size was larger. I also suspected that samp.dist would also be more distributed with more samples. 47 words Show parent | Reply Re: MATH1280 Unit 8 Discussion by Robert Hoot(Instructor) - Monday, 5 January 2015, 9:25 PM Hi Miguel, That was a great post. You were able to demonstrate what happens by creating a large sampling distribution and a large regular distribution. Sometimes for difficult statistical problems, you might not know the equation to answer the question with a paper and pencil, but you can simulate the situation and get a very good idea of the result. 60 words Show parent | Reply Re: MATH1280 Unit 8 Discussion by Miguel Cruz - Monday, 5 January 2015, 10:25 PM Thanks, Professor! 2 words
Show parent | Reply Re: MATH1280 Unit 8 Discussion by Nadezda Yudina - Tuesday, 6 January 2015, 9:30 AM Skillful use of R to simulate the large number of observations. 11 words Show parent | Reply Re: MATH1280 Unit 8 Discussion by Douglas Kibowen - Wednesday, 7 January 2015, 5:37 AM Good simulation. As the size increase the more the histogram for sample means tend towards the normal distribution. Good post. 20 words Show parent | Reply Re: MATH1280 Unit 8 Discussion by Whenu Adagba - Wednesday, 7 January 2015, 6:49 PM Impressive, this is really detailed. After running the simulation with rexp (15,3), I found that the distribution just remains almost the same and doesn't follow the normal approximations as I had expected. 34 words Show parent | Reply Re: MATH1280 Unit 8 Discussion by Ogwo Ifegwu - Wednesday, 7 January 2015, 11:40 PM Nice post, you have demonstrated a good of this distribution. I agree with the histogram will remain about this same as the all.obs histogram. 25 words
by Miguel Cruz - Monday, 5 January 2015, 11:50 AM Your suspicion was right. The "samp.dist" have a remarkable change, a more normal distribution to be more precise. Although, I use a million samples for "all.obs" histogram, the distribution was about the same as a single sample. 39 words Show parent | Reply Re: MATH1280 Unit 8 Discussion by Robert Hoot(Instructor) - Monday, 5 January 2015, 9:29 PM Hi Curtis, For the all.obs distribution, the numbers come from an exponential distribution (as shown in the example R code near the top of the discussion question). We can compare that to other exponential distributions and see that it is approximately in the expected shape. If we increased the number of samples, the shape might change a bit, but it would have the same general appearance of an exponential distribution (yes, the bars would be taller). For the other distribution, it does not seem to have a recognizable shape, but also notice that there are not many observations. If we obtained many more samples, then the central limit theorem would apply because that distribution measures the mean of the samples (it is a sampling distribution). 126 words Show parent | Reply Re: MATH1280 Unit 8 Discussion by Nadezda Yudina - Tuesday, 6 January 2015, 9:31 AM Good observations on the second histogram. I thought they all should distribute normally, but no. 15 words Show parent | Reply
Re: MATH1280 Unit 8 Discussion by Curtis Szmania - Thursday, 8 January 2015, 5:26 AM Thank you. It was an estimated guess! 7 words Show parent | Reply Re: MATH1280 Unit 8 Discussion by Douglas Kibowen - Wednesday, 7 January 2015, 5:39 AM Good discussion. when you increase the size the histogram of mean tends to be normal but for the all samples observations it seems to be unchanged. Good post. 28 words Show parent | Reply Re: MATH1280 Unit 8 Discussion by Whenu Adagba - Wednesday, 7 January 2015, 9:31 AM Hi, to be honest, I didn't even take notice of whether the bars became taller or not. I'd have to run this again and set out to monitor it 31 words Show parent | Reply Re: MATH1280 Unit 8 Discussion by Ogwo Ifegwu - Wednesday, 7 January 2015, 11:44 PM interesting observation, I was hoping that you will comment on the skewness of the histogram. 15 words Show parent | Reply Re: MATH1280 Unit 8 Discussion by Nadezda Yudina - Sunday, 4 January 2015, 7:15 PM
Show parent | Reply Re: MATH1280 Unit 8 Discussion by Oupnivesh Lollbeeharry - Wednesday, 7 January 2015, 1:56 AM Apparently the all.dist histogram for the 200 sample size does not look like a normal distribution. I thought it would look like a normal distribution same as you did. 30 words Show parent | Reply Re: MATH1280 Unit 8 Discussion by Whenu Adagba - Monday, 5 January 2015, 1:01 PM Question 1: What would you expect to happen to the histogram for samp.dist if we collected 200 more samples (from the same store) that each contain 15 observations? If we collected 200 more samples from the same store that each contain 15 observations, based on the Central Limit Theorem, I would expect that the distribution would be easily approximated using the Normal distribution which would produce a more meaningful histogram. In other words, the average of the sample distribution would be closer to the expectation, which means that it will be less spread. Running the simulation below X.bar <-rep (0,200) for (i in 1:200) { X.samp<- rexp(15,3) X.bar [i] <- mean (X.samp) } hist (X.bar), produces a histogram that is centred around the mean, 0.34 and is less spread. Using only 20 samples is a careless application of the Central Limit Theorem, because the sample size is small. Question 2: What would you expect to happen to the histogram for all.obs if we collected 200 more samples (from the same store) that each contain 15 observations? I would expect that collecting 200 more samples will reduce the spread of the graph, by about 10 times. As the sample size increases, the average of the distribution gets concentrated around the expectation. 100 words
Rate: Show parent | Reply Re: MATH1280 Unit 8 Discussion by Robert Hoot(Instructor) - Monday, 5 January 2015, 9:47 PM Hi Whenu, You proved your point using the simulation for the first part. On the second part, you said that the numbers would go closer to the mean, but those numbers come from an exponential distribution. We know from the book that exponential distributions have the shape similar to what we saw in the R code in the main discussion post. Try running this code and see what you get when you take 200 samples of size 15 from an exponential distribution:
Create an array to hold 200 rows of data
that has 15 columns each (200 samples of n=15):
X.samples <- array(rep(0, 200*15), dim=c(200,15))
Generate 200 samples of n=15 from an exponential
distribution:
for (i in 1:200) { X.samples[i, ] <- rexp(15, 3) } hist (X.samples) 138 words Show parent | Reply Re: MATH1280 Unit 8 Discussion by Whenu Adagba - Wednesday, 7 January 2015, 6:45 PM I finally did run with 200, then 20000 and then 2000000. The histogram practically does not converge to the expectation as the number of samples selected increases like I earlier posted. I guess I was wrong then. 37 words Show parent | Reply
