Modeling data distributions #4 - Questions and Answers

Question 1 A standard 3σα chart has been enhanced with early warning limits at plus-minus one o from the centerline. Three sample means in a row have plotted above the +10 line. what is the probability of this happening if the process is still in control. Answer Introduction: The empirical rule for a normally distributed random variable is as follows: About 68% of the data lie within the 1-standard deviation interval about the mean, About 95% of the data lie within the 2-standard deviation interval about the mean, About 99.7% of the data lie within the 3-standard deviation interval about the mean. Explanation: When the process is still in control, the sample mean can be said to follow the normal distribution with the true value of the characteristic of interest as the mean (center line), and the relevant standard error of sample mean. In this case, the above empirical rule will hold true for the process. Calculation: Assume X1, X2, X3 to be three consecutive sample means obtained from the process. Since the samples are all obtained at random, the values taken by the sample mean will be independent of one another. Further, being taken from the same process, X1, X2, X3 are identically distributed. Now, it is given that each of the three means are above the +1-standard deviation line, that is, above the upper bound of the 1-standard deviation interval of the mean. Consider the case of X1. Since 68% of the data lie within the 1-standard deviation interval, the remaining 32%(= 100%-68%) lie outside the 1-standard deviation interval. The normal distribution being symmetric, half of this 32% would lie above the upper bound of the 1-standard deviation interval, while the remaining half would lie below the lower bound of the interval. Thus, X₁ would lie above the upper bound of the 1-standard deviation interval, that is, above the +1-standard deviation line, with a 16%(= 32%/2) or 0.16 probability. Since, X1, X2, X3 are identically distributed, each of bar X2 and X3 would lie above the +1-standard deviation line with probability 0.16. The distributions of X1, X2, X3 being independent, the probability that all three of bar X1, X2, X3 lie above the +1-standard deviation line is the product of the individual probabilities, that is, (0.16) (0.16) (0.16) = 0.004096. Hence, if the process is still in control, the probability that 3 sample means in a row would plot above the +1-standard deviation line is 0.004096. Question 2 The admissions office of a college wants to determine whether there is a relationship between IQ scores x and grade-point averages y after the first year of school. An equation that models the data obtained by the admissions office is y = 0.069x-4.755. Estimate the values of x that predict a grade-point average of at least 3.5. (Simplify your answer completely. Round your answer to the nearest whole number.) Answer The given equation modeling the relationship between IQ score and grade point averages is as follows y = 0.069x-4.755... (1) Where y represents grade point averages And x represents IQ score We need to find value of x for y = 3.5 So substituting y = 3.5 in equation (1), we get 3.5 0.069x - 4.755 On simplifying the above equation 0.0692 3.5 +4.755 0.069x8.255 8.255 x= 0.069 x = 119.63 So the next whole number is 120. Hence for IQ score that is x = 120, the grade point average will be at least 3.5. Answer: For grade point average of at least 3.5 required IQ score is 120 Question 3 Learning math The Core Plus Mathematics Project (CPMP) is an innovative approach to teaching Math- ematics that engages students in group investigations and mathematical modeling. After field tests in 36 high schools over a three-year period, researchers compared the performances of CPMP students with those taught using a traditional curriculum. In one test, students had to solve applied Algebra problems using calculators. Scores for 320 CPMP students were compared to those of a control group of 273 students in a traditional Math pro- gram. Computer software was used to create a confidence interval for the difference in mean scores. (Journal for Research in Mathematics Education, 31, no. 3[2000]) Conf level: 95% Variable: Mu (CPMP) - Mu(Ctrl) Interval: (5.573, 11.427) a) What's the margin of error for this confidence interval? b) If we had created a 98% CI, would the margin of error be larger or smaller?c) Explain what the calculated interval means in context. d) Does this result suggest that students who learn Mathematics with CPMP will have significantly higher mean scores in Algebra than those in traditional programs? Explain. Answer (a) The margin of error can be calculated as: Marg E oferr or Upper lim it - Lower lim it 11.427-5.573 2 = 2.927 2 (b) The margin of error is positively related to the confidence level. The increase in confidence level tends to increase the margin of error and vice versa. So, it could be said that the margin of error would be larger for 98% confidence level. (c) There is 95% probability that the difference in true mean score of student of CPMP and those of traditional curriculum lies between 5.573 and 11.427. Question 4 A standard 3 sigma x-bar chart has been enhanced with early warning limits at plus- minus one sigma from the centerline. Three sample means in a row have plotted above the +1 sigma line. what is the probability of this happening if the process is still in control. Answer Step 1 Introduction: The empirical rule for a normally distributed random variable is as follows: About 68% of the data lie within the 1-standard deviation interval about the mean; About 95% of the data lie within the 2-standard deviation interval about the mean; About 99.7% of the data lie within the 3-standard deviation interval about the mean. Step 2 Explanation: When the process is still in control, the sample mean can be said to follow the normal distribution with the true value of the characteristic of interest as the mean (center line), and the relevant standard error of sample mean. In this case, the above empirical rule will hold true for the process. Step 3 Calculation: Assume X1, X2, X3 to be three consecutive sample means obtained from the process. Since the samples are all obtained at random, the values taken by the sample mean will be independent of one another. Further, being taken from the same process, X1, X2, X3 are identically distributed. Now, it is given that each of the three means are above the +1-standard deviation line, that is, above the upper bound of the 1-standard deviation interval of the mean. Consider the case of X1 Since 68% of the data lie within the 1-standard deviation interval, the remaining 32% = (100% - 68%) lie outside the 1-standard deviation interval. The normal distribution being symmetric, half of this 32% would lie above the upper bound of the 1-standard deviation interval, while the remaining half would lie below the lower bound of the interval. Thus, X₁ would lie above the upper bound of the 1-standard deviation interval, that is, above the +1-standard deviation line, with a 16% = (32%) or 0.16 probability. Since X1, X2, X3 are identically distributed, each of X₂ and X3 would lie above the +1- standard deviation line with probability 0.16. The distributions of X1, X2, X3 being independent, the probability that all three of X1, X2, X3 lie above the +1-standard deviation line is the product of the individual probabilities, that is, (0.16) (0.16) (0.16) - 0.004096 Hence, if the process is still in control, the probability that 3 sample means in a row would plot above the +1-standard deviation line is 0.004096. Question 5 1.What factors influence the correspondence between the binomial and normal distributions? 2.Twenty percent of individuals who seek psychotherapy will recover from their symptoms irrespective of whether they receive treatment. A research finds that a particular type of psychotherapy is successful with 30 out of 100 clients. Using an alpha level of 0.05 as a criterion, what should she conclude about the effectiveness of this psychotherapeutic approach? 3.How does the size of the data set help cut down on the size of the error terms in the approximation process? Answer Step 1 Note: Hey, since multiple questions are posted, we will answer first question for you. If you want any specific question to be answered, then please submit that question only or specify the question number in your message. Step 2 1. Binomial approximation: The normal distribution can be used as an approximation to the binomial distribution, under certain circumstances, namely: If X ~ B (n, p) and if n is large and/or p is close to 1/2, then X is approximately N (np, npq) (where q = 1 - p). Step 3 Continuity Correction: The binomial is a discrete random variables, whereas the normal distribution is continuous. We need to take this into account when we are using the normal distribution to approximate a binomial using a continuity correction. Continuity correction factor table: If P(X = n) use P(n-0.5n) use P(X>n +0.5) If P(X < n) use P(X n) use P(X>n-0.5) Continuous normal distribution can be used as an approximation to binomial distribution, the modification is known as continuity correction.