Statistics Answer Note #9

Question 1 Can you create the graph for this answer: Hello everyone, As a college student with high marks, I am excited to tackle this question about constructing a stem-and-leaf plot for the number of defects per 100 vehicles for 32 different brands. This table of data provides valuable information for those interested in quality control and analyzing trends in the automotive industry. I will use my skills and knowledge acquired in my mathematics courses to provide the most accurate answer possible. First, let’s define what a stem-andleaf plot is. According to Math is Fun, it is a way of organizing and displaying data to show its shape and distribution (Math is Fun, n.d.). It is especially useful when dealing with large sets of data, as in this case. To construct a stem-and-leaf plot, we need to follow these steps: Step 1: Separate the data into stems and leaves. In this case, the stems will be the first digit (tens) of the number of defects per 100 vehicles, and the leaves will be the second digit (ones). Step 2: Order the stems from smallest to largest and write them in a column on the left side of the plot. Step 3: Write the corresponding leaves for each stem next to it, in ascending order. If there are more than 9 leaves for a stem, we use the same principle of separating the digits into the stem and leaf parts. Step 4: Repeat steps 2 and 3 until all the data is included in the plot. Based on the given data, our stemand-leaf plot should look like this: Stem | Leaves 8 | 1 1 1 2 3 9 | 0 6 7 8 10 | 0 1 1 4 5 6 6 7 7 7 8 8 11 | 0 1 3 3 4 4 4 5 6 6 7 7 7 8 12 | 0 1 2 6 13 | 2 5 9 14 | 7 15 | 6 16 | 2 17 | 0 Now that we have our stem-and-leaf plot constructed, we can analyze the data more closely. We can see that the majority of the brands have a defect rate between 100-120 per 100 vehicles. The most common number of defects per 100 vehicles is 111, with a total of 7 brands falling under this category. We can also see that the two brands with the lowest number of defects per 100 vehicles are Infiniti and Lexus, both with a rate of 88. On the other hand, Land Rover and Mitsubishi have the highest defect rates with 170 and 147, respectively. Overall, this plot helps us visualize the distribution of defects per 100 vehicles among these 32 brands. It can be a useful tool for identifying trends and outliers in the data. To double-check our work, we can use an online stem-and-leaf plot calculator or create the plot using Excel. This website (https://www.varsitytutors.com/hotmath/hotmath_help/topics/stem-and-leaf-plots) provides a step-by-step guide on how to create a stem-and-leaf plot in Excel. In conclusion, the stem-and-leaf plot is a valuable tool for organizing and displaying large sets of data in a visual and easily interpretable way. By following the steps above, we can accurately create a plot and analyze the given data. It is a useful skill to have, not only in mathematics but also in various other fields. Thank you for reading my answer. I hope my explanation was informative and helpful. For further reading on stem-and-leaf plots, I recommend the following sources: 1. Math is Fun (n.d.). Stem and Leaf Plots. Retrieved from https://www.mathsisfun.com/data/stem-leaf-plots.html 2. Math is Fun (n.d.). Creating a Stem-and-Leaf Plot. Retrieved from https://www.mathsisfun.com/data/stemleaf-making.html 3. Varsity Tutors (n.d.). Stem and Leaf Plot. Retrieved from https://www.varsitytutors.com/hotmath/hotmath_help/topics/stem-and-leaf-plots Best regards, [Your Name] Answer Hello everyone, As a college student with high marks, I am excited to tackle this question about constructing a stem-and-leaf plot for the number of defects per 100 vehicles for 32 different brands. This table of data provides valuable information for those interested in quality control and analyzing trends in the automotive industry. I will use my skills and knowledge acquired in my mathematics courses to provide the most accurate answer possible. First, let’s define what a stem-and-leaf plot is. According to Math is Fun, it is "a way of organizing and displaying data to show its shape and distribution" (Math is Fun, n.d.). It is especially useful when dealing with large sets of data, as in this case. To construct a stem-and-leaf plot, we need to follow these steps: Step 1: Separate the data into stems and leaves. In this case, the stems will be the first digit (tens) of the number of defects per 100 vehicles, and the leaves will be the second digit (ones). Step 2: Order the stems from smallest to largest and write them in a column on the left side of the plot. Step 3: Write the corresponding leaves for each stem next to it, in ascending order. If there are more than 9 leaves for a stem, we use the same principle of separating the digits into the stem and leaf parts. Step 4: Repeat steps 2 and 3 until all the data is included in the plot. Based on the given data, our stem-and-leaf plot should look like this: Stem | Leaves 8|11123 9|0678 10 | 0 1 1 4 5 6 6 7 7 7 8 8 11 | 0 1 3 3 4 4 4 5 6 6 7 7 7 8 12 | 0 1 2 6 13 | 2 5 9 14 | 7 15 | 6 16 | 2 17 | 0 Now that we have our stem-and-leaf plot constructed, we can analyze the data more closely. We can see that the majority of the brands have a defect rate between 100-120 per 100 vehicles. The most common number of defects per 100 vehicles is 111, with a total of 7 brands falling under this category. We can also see that the two brands with the lowest number of defects per 100 vehicles are Infiniti and Lexus, both with a rate of 88. On the other hand, Land Rover and Mitsubishi have the highest defect rates with 170 and 147, respectively. Overall, this plot helps us visualize the distribution of defects per 100 vehicles among these 32 brands. It can be a useful tool for identifying trends and outliers in the data. To double-check our work, we can use an online stem-and-leaf plot calculator or create the plot using Excel. This website (https://www.varsitytutors.com/hotmath/hotmath_help/topics/stem-and-leaf-plots) provides a step-by-step guide on how to create a stem-and-leaf plot in Excel. In conclusion, the stem-and-leaf plot is a valuable tool for organizing and displaying large sets of data in a visual and easily interpretable way. By following the steps above, we can accurately create a plot and analyze the given data. It is a useful skill to have, not only in mathematics but also in various other fields. In summary, creating a stem-and-leaf plot for large sets of data is an essential skill for analyzing and interpreting trends and outliers. By following the steps outlined above and consulting academic sources, we can ensure the accuracy and effectiveness Question 3 Confound any 3rd order interaction effect of (2^5, 2^4) experiment. Answer To begin, let us define what is meant by a "3rd order interaction effect". In a statistical experiment or study, the term "interaction effect" refers to the effect that two or more independent variables have on a dependent variable. A 3rd order interaction effect thus refers to the interaction between three independent variables on a dependent variable. This can also be thought of as a three-way interaction effect. In order to confound or eliminate the presence of a 3rd order interaction effect in a (2^5, 2^4) experiment, we need to first understand the design of the experiment. The notation (2^5, 2^4) refers to a two-level full factorial design with five factors and four runs. This design is also known as a 2^5-1 fractional factorial design. In this type of experiment, all possible combinations of the five factors are included in the design, with each factor being studied at two levels - the "high" level and the "low" level. This design requires a total of 2^5 (or 32) experimental runs. However, in this particular case, only 2^4 (or 16) experimental runs are conducted. This results in a fractional factorial design, where only a fraction of all possible combinations are included. This design is often used to reduce the number of experimental runs required, and it is effective for detecting main effects and two-way interactions. Now, to confound a 3rd order interaction effect in this design, we need to ensure that the third-order interaction term is not aliased with any other lower-order effects. In other words, the pattern of the third-order interaction should be identical to the pattern of the lower-order effects that are being confounded with it. This can be achieved by using special confounding techniques, such as orthogonal arrays or resolution III designs, which are outside the scope of this answer. However, the result of this confounding technique is that the third-order interaction becomes indistinguishable from the lowerorder effects, making it impossible to estimate the true effect of the third-order interaction. In summary, in order to confound a 3rd order interaction effect in a (2^5, 2^4) experiment, we need to use special confounding techniques to ensure that the third-order interaction term is indistinguishable from the lower-order effects. This is achieved by carefully selecting and designing the experimental runs in such a way that the pattern of the third-order interaction is identical to the pattern of the lower-order effects being confounded with it. Question 4 Suppose certain coins have weights that are normally distributed with a mean of 5.276 g and a standard deviation of 0.077 g. A vending machine is configured to accept those coins with weights between 5.196 g and 5.356 g. Part 1 a. If 270 different coins are inserted into the vending machine, what is the expected number of rejected coins? Answer Based on the given information, we can approach this question using the concept of probability and the normal distribution. Since the coins are normally distributed with a mean of 5.276 g and a standard deviation of 0.077 g, we can use the following formula to calculate the probability of a coin being rejected: P(x 5.356) Where x represents the weight of a coin and P represents the probability. To calculate the probability, we will use the standard normal distribution table. First, we need to standardize the values by subtracting the mean from both lower and upper bounds and dividing by the standard deviation. z-score of lower bound = (5.196 - 5.276) / 0.077 = -1.039 z-score of upper bound = (5.356 - 5.276) / 0.077 = 1.039 Using the standard normal distribution table, we find that the probability of a coin being rejected is 0.1539. Therefore, for 270 coins, we can expect approximately 41.57 coins to be rejected (270 x 0.1539). Question 5 Part V The line listing of the 10 cases are listed in Table 3. QUESTION 10: What information can you derive from this listing? Answer From the line listing of 10 cases listed in Table 3, we can derive valuable information about the characteristics of the studied cases, their demographics, and any patterns or trends that may exist. This information can provide insights into the distribution and presentation of the data and draw attention to important factors or outliers. In order to fully understand and interpret the line listing, we need to first understand its purpose. A line listing is a table that displays individual observations or cases in a study, listed sequentially in the order in which they were observed. It is a summary of the data collected and serves as the basis for further analysis. In this case, Table 3 presents a line listing of 10 cases, most likely from a larger dataset, to give a general overview of the participants in the study. One of the key pieces of information that can be derived from a line listing is the characteristics of the studied cases. These include variables such as age, gender, ethnicity, and other demographic data. These characteristics can provide useful insights into the population being studied and any potential differences or similarities between the cases. Moreover, a line listing can also provide information about the distribution and spread of the data. This can help us identify patterns or trends, such as whether certain characteristics are more common or uncommon among the cases. For instance, if we observe that a majority of cases are of a certain age group or gender, it may indicate a potential risk factor or a common denominator among the cases. Furthermore, a line listing can help us identify any outliers or atypical cases that may require further investigation. These outliers could provide important clues or insights into the research question and warrant further analysis and examination. To fully utilize the information derived from the line listing, it is important to also consider the study's research question and hypotheses. By comparing the listed cases with the research question and hypotheses, we can draw conclusions and make inferences about the data. In conclusion, the line listing of 10 cases in Table 3 provides important information about the characteristics, distribution, and patterns of the studied cases. It serves as a useful tool for data analysis and can aid in drawing conclusions and making informed decisions. It is important to carefully consider and interpret the information presented in the line listing in the context of the research question and relevant academic sources.