**Observations on Degree of Polymerization**An article provided the following observations on the degree of polymerization for paper specimens, where viscosity times concentration fell in a specific middle range:- Observations: 415, 420, 421, 422, 426, 427, 432, 434, 437, 438, 446, 447, 450, 452, 457, 463, 464**(a) Boxplot Construction**A set of four boxplots is depicted, each representing the same data. The boxplots show the distribution of observations in terms of their quartiles, median, and potential outliers.**Features to Comment On:**- There is one outlier.- The data is strongly skewed.- The data appears to be centered near 428.- There are no outliers.- The data appears to be centered near 438.- There is little or no skew.**(b) Evaluation of Normal Distribution**- Is it plausible that the sample observations were selected from a normal distribution? - Options: Yes / No**(c) Confidence Interval Calculation**- Task: Calculate a two-sided 95% confidence interval for the true average degree of polymerization. Answers should be rounded to two decimal places.**Questions:**1. Does the interval suggest that 440 is a plausible value for the true average degree of polymerization? - Options: Yes / No2. Does the interval suggest that 456 is a plausible value? - Options: Yes / No**Graph and Diagram Explanation:**The boxplots display the central tendency, variability, and potential outliers within the data set. Each plot visually represents the quartiles and the overall spread of the data, allowing for interpretation of skewness and the presence of any anomalies.

(a) Use the MINITAB to draw the boxplot for degree: Procedure: Choose Graph > Boxplot. Under One Y, choose Simple. Click OK. In Graph variables, enter the data of Degree. In Scale, select Transpose value and Category scales under Axes and Ticks. Click OK. Output: Correct answer: Option 4 From the boxplot, it can be observed that, all the data points are centered around the value 428 (middle line of boxplot-median). There are no outliers in the sample. The data is very slightly left skewed because middle line of boxplot is very slightly closer to left bottom when compared to right. Correct answer: There are no outliers. The data appears to be centered near 438. There is little or no skew. (b) From the boxplot, it can be observed that, the distribution is very slightly left skewed indicating that the distribution might be approximately normally distributed. Since the sample is approximately normal, it might be plausible that observations might have come from normal distribution. Correct answer: Yes(c) The descriptive statistics is conducted using EXCEL. The software procedure is given below: Enter the data. Select Data > Data Analysis >Descriptive Statistics> OK. Enter Input Range as degree column. Mark Labels in First Row. Mark Summary Statistics. Click OK. The output using EXCEL is as follows: From the output, the mean is 438.29, and the standard deviation is 15.5353. The degrees of freedom is, df=n-1=17-1=16 The degrees of freedom is 16. The confidence level is 95%, than the level of significance is 0.05. Computation of critical value: The critical value of t-distribution for 0.05 level of significance at 16 degrees of freedom can be obtained using the excel formula “=T.INV.2T(0.05,16)”. The critical value is 2.1199. The 95% confidence interval for true average degree of polymerization is, CI=x¯±tα2sn=438.29±2.119915.535317=438.29±7.99=438.29-7.99,438.29+7.99=430.30,446.28 Thus, the 95% confidence interval for true average degree of polymerization is (430.30,446.28). Since the confidence interval contains the value 440, the interval suggest that 440 is a plausible value for true average. Correct answer: Yes The confidence interval does not contain the value 456, the interval does not suggest that 456 is a plausible value for true average. Correct answer: No

Math

Statistics

(a) Construct a boxplot of the data. 420 430 440 450 460 420 430 440 450 420 430 440 450 460 420 430 440 450 460 Comment on any interesting features. (Select all that apply.) There is one outlier. The data is strongly skewed. The data appears to be centered near 428. There are no outliers. The data appears to be centered near 438. There is little or no skew. (b) Is it plausible that the given sample observations were selected from a normal distribution? Yes No (c) Calculate a two-sided 95% confidence interval for true average degree of polymerization. (Round your answers to two decimal places.) Does the interval suggest that 440 is a plausible value for true average degree of polymerization? Yes No Does the interval suggest that 456 is a plausible value? Yes No

(a) Construct a boxplot of the data. 420 430 440 450 460 420 430 440 450 420 430 440 450 460 420 430 440 450 460 Comment on any interesting features. (Select all that apply.) There is one outlier. The data is strongly skewed. The data appears to be centered near 428. There are no outliers. The data appears to be centered near 438. There is little or no skew. (b) Is it plausible that the given sample observations were selected from a normal distribution? Yes No (c) Calculate a two-sided 95% confidence interval for true average degree of polymerization. (Round your answers to two decimal places.) Does the interval suggest that 440 is a plausible value for true average degree of polymerization? Yes No Does the interval suggest that 456 is a plausible value? Yes No

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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