Statistics Final Exam Analysis - Assignment Example

Take home final Exam Question a) Female graph representing the relationship between height and arm Male graphrepresenting the relationship between height and arm Female height-armspans for many females are equal. Many males had different height-armspans b) Ho: The distance from the sole of the feet to the tip of the head is exactly the same as the distance between the outstretched hands. Ha: The distance from the sole of the feet to the tip of the head is not exactly the same as the distance between the outstretched hands c) Null hypothesis almost mirrored the result of the female comparison. Alternative hypothesis reflect the result of male distance disparities and part of female disparity. d) N= 272 (∑X) =18059 ∑Y = 17808 ∑ XY = 1184126 ∑X2 = 1200851 ∑Y2 1168864 (∑Y)2 = 317124864 (∑X)2 326127481 r = 272(1184126) – (18059*17808) /√ {272(1200851) – (326127481)}* {√272 (1168864)- (317124864)} = 487600/ (709.923*897.855) =0.765 Data is not skewed e) NO. There is no significant relationship between the variables tabulated in the data. Notably, plotting the graph of height against arm does not constitute to a line of best fit. The only way to make the data to adapt to the linear regression is to introduce coefficient of correlation. f) If r = 0.765, then r 2 = 0.585, which means that 59% of the total variation in y can be explained by the linear relationship between x and y. The other 41% of the total variation in y remains unexplained. g) From the diagnostic plots, what can you say about your analysis? i.e. is what you did the appropriate analysis? (1pt) No: diagnostic plots could not be achieved h) Transform the arm variable by taking its reciprocal, then carryout the regression analysis again. Report all your results. Compare the r-squared in this analysis to the one in the previous analysis if it is bigger then, the new analysis is much better than the previous. Is this the case based on the diagnostic plot? (1pt) Reciprocal values are less Question 2 a. The table below gives the mean selling price for the Palm M515 PDA over a one-week selling period in May 2003 on amazon.com. Sample size Sample Mean Standard Deviation (s) 18 231.611 21.936 i. What is the margin of error for a 90% confidence interval? (2pts) n =18 mean = 231.611 sd = 21.936 x-bar ± (t critical value) * s/SQRT(n) 231.611 ± 1.7 *21.936/√18 =389.824 ii. What are the degrees of freedom? (1pt) df sample size = 18-1 = 17 df sample mean = 231.611 – 1 = 230.611 iii. Construct a 90% confidence interval for the mean selling price, and do not forget to interpret your interval. (2pts) b. i. Can this be considered binomial? If yes, list the properties that make it binomial. (1pt) Yes: Properties of binomial in the question It has n fixed identical trials Its trials are independent P(S) =p and P (F) =q where q = 1−p. Each variable has a success and ii. Define the random variable X. (1pt) The experiments are yes no in nature (yes it is or no it is not). And there is a probability (p) that any given number is bad. iii. What is the probability that exactly 3 of the Millennial are unaffiliated to any faith? (1pt) The probability that 3 of the millennial are unaffiliated is 3/9 =1/3 iv. What is the mean number of Millennial unaffiliated to any faith? (1pt) Mean number n*p = 5*.1 = 9*1/3 =3 v. What is the standard deviation? (1pt) sd =sq Question 3 1. In a state lottery (scratch-card game), a player wins $50 by matching the given symbol from 3 different rows, and wins $10 by matching the given symbol in two different rows else they do not win anything. If the first row contains 10 place holders, the second row contains 7 place holders, and the last row contains 5 place holders. Of the place holders in rows only one has the given symbol, and a player is allowed to scratch only one place holder in each row else the play is invalid. (Hint: A player is only interested in the number that they match to know their winnings). a. Define the random variable X, and construct a probability distribution table for the random variable X. (3pts) Then X is a random variable whose realm is the sample space and whose range is the set 10; 7; and 5: X can be interpreted as the gain of a player in a game in which a card is picked and matched, the player winning $50 if the outcome is matches the three rows of 10, 7, 5 and winning $10 if the outcome matches 2 rows and wins nothing if no matches. The probability diagram Win $ 50 Win $ 10 Win $0 probability 3/22 1/11 0 b. What is the probability that someone matches at least 2? (1pt) Probability of matching two is 2/22 c. Find the Expected value and the standard deviation of X. (2pts) Variance 10 – 7 = 3 7 – 5 = 2 d. Explain the meaning of the expected value of X calculated above. (1pt) e. If it costs $2 to buy a scratch-card and on average about 100,000 cards are sold in a day, find how much the state makes in a day by selling the scratch-card game. (3pts) Cost Question 4 a. The result of the study could not be legitimate since the participants were not randomly picked and were selected from known source. The study is an ethical since research should not be coaxed and should be a voluntary participation event. The facilitation fee that was given by legal aid board is unethical and made study illegitimate. b. Variance is the difference from the sample data. The higher the difference the higher the probability that estimation would be less accurate. The more efficient estimator would be the one with variance of 2. Compared to variance of 5 c. Mean =15 sd= 6 Size = 9 The central limit theorem (CLT) states that, given assured conditions, the mean of a adequately large figure of iterates of independent random elements, each with a well-started mean and well-illustrated variance, will be normally distributed (Fischer 187). That is, assume that a sample is gotten containing a large number of surveys, each survey being randomly created in a way that does not rely on the values of the other surveys, and that the calculation average of the observed elements is computed. d. The sample should be almost to equal actual population size to give a small error margin and a 95% confidence. Change on the sample size would increase accuracy and reduce error margin e. Simple Random Sampling This is the most commonly known category of a random sample. This is typified by the fact that the probability of sampling is the same for every circumstance in the population. Simple random sampling is a technique of selecting n units from a sample population of size N such that every likely sample of size a has equal probability of being drawn. Stratified Random Sampling In this type of sampling, the sample population is first split into two or more mutually exclusive sections based on some groups of variables of interest in the study. It is designed to systematize the population into homogenous segments before sampling, then drawing a random sample in each subset. With this method the population of N units is divided into sub sizes of units respectively. These subpopulations, known as strata, are non-overlapping and mutually they comprise the entire population. When these have been established, a sample is picked from each, with a different draw for each of the varied strata. The sample sizes in the strata are indicated by respectively. If a SRS is picked within each stratum, then the whole sampling process is referred to as stratified random sampling. Systematic Sampling This technique of sampling is at first glance very unlike SRS. In practice, it is an alternative of simple random sampling that engages some listing of elements - every nth variable of list is then picked for inclusion in the sample. For example, a list of 10,000 people and require a sample of 1,000. Creating such a sample includes three steps: Divide number of scenarios in the population by the appropriate sample size. In this case, dividing 10,000 by 1,000 top get a value of 10. Select a random figure between one and the value obtained in Step 1. In this case, choose a figure between 1 and 10 – such as 7. Beginning with case figure chosen in Step 2, pick every tenth record Question 5 AU student wanted to know if Caribou coffee charges more than Starbucks for their drinks. The student did a random sample of drink prices from both shops at different locations in the Washington DC area to compare the two coffee shops. Using the methods discussed in class answer the following questions? (Load data from statcrunch.com file name Starbucks/Caribou.) a. Do graphical summaries on the data, and give a short summary of your observations. (2pts) b. Is it appropriate to carryout significance test on the data? Give reasons. (2pts) c. If the answer to the above is yes, state the null and alternative hypotheses. (1pt) d. Give the results of your significance test as displayed in Statcrunch, and state your decision and conclusion. (3pts) e. How is the p-value given in the results calculated? (1pt) f. In your results there is a value called standard error explain what it is and how it is found? (1pt) Question 6 2. The North Carolina department of game and fish tracks the size of fish in its rivers. Two rivers (Wacamaw and Lumber) were selected and through a random sample of fish they found that the Wacamaw River fish were on average 40.39cm long with a variance of 75.69, and the Lumber River fish were on average 39.41cm long with a variance of 69.06. Assuming the distribution of fish length is normal in both rivers find: a. The percent of fish in the Wacamaw River that are longer than 51.7cm (2pts) Average length = 40.39cm b. The percent of fish in the Lumber River that are shorter than 19.1cm (2pts) c. What percent of fish in the Wacamaw River fall between 20cm and 51.7cm in length? (3pts) d. What length of fish in the Lumber River corresponds to the lower 17%? (3pts) e. Using the 68-95-99% rule, find the lengths of fish in the middle 95% in both rivers. Which one has the wider range? What contributes to this? (3pts) Work Cited Fischer, Hans. A History of the Central Limit Theorem: From Classical to Modern Probability Theory. New York: Springer, 2011. Print. Read More