**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

See similar textbooks

Similar questions

How do the medians compare between the two dot plots
Use a stem-and-leaf plot to display the data, which represent the numbers of hours 24 nurses work per week. Describe any patterns. 33 40 36 53 32 36 40 35 40 40 41 48 30 26 40 36 40 36 40 33 40 32 38 29 Determine the leaves in the stem-and-leaf plot below. Key: 313 33 Hours worked 3 4
Use the same scales to construct modified boxplots for the pulse rates of males and females from the accompanying data sets. Identify any outliers. Use the boxplots to compare the two data sets. Click the icon to view the data sets. Determine the boxplot for the men's pulse rate data. Choose the correct graph below. O A. O C. xx HE 40 50 60 70 80 90 100110 O C. What points are outliers? OA. 46 O B. 46, 89, 105 OC. 46, 48, 105 O D. 105 40 50 60 70 80 90 100110 -ידי xH Determine the boxplot for the women's boxplot data. Choose the correct graph below. O A. Q 40 50 60 70 80 90 100110 xH Q 40 50 60 70 80 90 100110 Q ✔ Q ✔ C O B. O D. O B. O D. 841x 40 50 60 70 80 90 100110 X 40 50 60 70 80 90 100110 40 50 60 70 80 90 100110 XX 40 50 60 70 80 90 100110 Q Q M Q Q ✔ Pulse rates for men and women Men's Pulse Rates 59 72 72 58 84 50 58 65 66 55 105 64 65 77 70 62 Women's Pulse Rates 76 70 74 83 59 67 65 72 74 89 85 78 56 80 60 46 70 59 70 75 66 68 66 72 77 78 97 74 62 81 54 74 77 79 89 95 59 98…
The data to the right shows the number of graduating students with business majors at a university over the last 15 years. 277 269 263 256 247 241 236 230 224 214 208 202 a. Construct a box-and-whisker plot for this data. b. Identify the five-number summary. 197 187 182 斯 a) Choose the correct box-and-whisker plot of the data below. C. O B. OA. 150 200 250 300 150 200 250 300 150 200 250 300 b) The five-number summary is (Use ascending order.)
Please help with A and B, Also, does this data set contain any outliers? Yes or No
Please answer the question in the photo. TY!
Please draw the box plot for the given data.
Use a stem-and-leaf plot to display the data, which represent the scores of a biology class on a midterm exam. Describe any patterns. 75 91 75 80 85 90 80 92 94 68 87 68 82 88 95 75 87 91 91 87 76 91 85 99 (...)) Determine the leaves in the stem-and-leaf plot below. Key: 313=33 Exam Scores 9 What best describes the data? OA. Most of the scores are in the 70s and 80s. OB. Most of the scores are in the 60s and 70s. OC. Most of the scores are in the 80s and 90s. OD. The scores are evenly spread from the 60s to the 90s.
Find the 5 number summary for the data shown 3 18 24 26 28 38 55 63 65 67 5 number summary: 3 65 64 64 X, 21 IQR:|| The 1.5 x IQR rule states that values between and are likely not outliers.
A scatter diagram can be used to visualize a pattern to the data. O True O False
Show work and appropriate explanation please.