Business Statistics Homework 18 Answesr

CHAPTER 1 EXERCISES 1. For each of the variables below, compute by hand (that is, with a calculator) their mean and standard deviation. Confirm your results with Minitab. (Type the numbers into separate columns in the Minitab worksheet.) - For variables a, b, and c, what relationships do you see between the variable means? Between the variable standard deviations? - Stat 225 Review for Extra credit (3 points): Note that Variable e = Variable b + Variable d. Note that the standard deviation of e ≠ sum of standard deviations of b and d? What is the correct relationship? (Consider this a research question. You are not expected to derive the formula mathematically.) Variab le a. 100, 200, 300, 400, 500, 600, 700, 800, 900 Mea n A Variab le b. 1, 2, 3, 4, 5, 6, 7, 8, 9 Mea n B Variab le c. 90, 190, 290, 390, 490, 590, 690, 790, 89 C d. 5, 8, 13, 21, 34, 55, 89, 144, 233 e. 6, 10, 16, 25, 39, 61, 96, 152, 242 StDe v 500.0 273.9 StDe v 5.000 2.739 Mea n StDe v 490.0 273.9 Variab Mea StDe le n v D Variab Mea66.9StDe77.1 le n v E 71.9 79.6 2. a. For 2021-22, the minimum NBA player salary is $925,000. The mean salary is approximately $8.1 million, the median $4.0 million, and, at the time of this writing, the maximum salary is $46 million (Steph Curry). Do you believe the distribution of 2020-21 NBA player salaries is a normal distribution? Briefly explain your answer. No this is not a normal distribution; the income is skewed because of the maximum salary seems to be an outlier. This is also shown with the significant difference in the mean and the median. b. Consider the distribution of all household incomes in the state of Indiana. Briefly explain why you believe this is not a normal distribution. The distribution of all household incomes in the state of Indiana is positively skewed with a long right tail. The mean would be to the right of the median. The number of people making salaries this large is very low while the number of people with low/average incomes is very high. 3. a. Open the “How Much is a Win Worth” document. Evaluate this graph (which is from the Spring 2018 Exponent). Give two criticisms of the graph. 1.The circles are different sizes for no apparent reason which makes the graph confusing. 2. There is no X axis b. Open the UK 2016 Q4 GDP Growth graph. What is a major flaw in its construction? (Hint: Is the Second Estimate of the GDP growth three times bigger than the First Estimate?) Because of the short range of the graph the 10% difference in the estimates appears much larger than it actually is. 4. Suppose the Pacers basketball team is averaging 108 points per game half way through their season. a. Is this a good prediction of exactly how many points they will score in their next game? Why or why not? No because it is only an average so we cannot see how many times they scored around this many points, way above, or below. b. To make such a prediction, what other information would you like to have? The teams scores for each game rather than just an average. 5. In an upcoming basketball game, New Orleans at Utah, sports gamblers can bet on whether Zion Williamson, a player for New Orleans, will score more or fewer than 26.5 points. An analyst points out that in his 4 career games against Utah last year, Zion is averaging 23.5 points per game. What would you rather know? A graph showing how many games he scores over or under his average. - Extra Credit (2 pts): Briefly explain the following statement using the probability and mathematical concepts covered in this chapter: “If you focus on means and differences of means, you will make a lot of mistakes about individuals.” – James Q. Wilson, famous social scientist. The mean or medians are not a good indicator of data because there can be outliers. 6. Tracy, an Indiana high school student, is considering applying to Purdue but notices the 16% admission rate for all applicants. What conditional probabilities should Tracy look at to get a better picture of the applicable acceptance rate? (Give two.) The average GPA of the accepted students and the number of applicants the school gets each year. 7. Read the two documents in the Washington State Climate Change Results file. Briefly explain how response bias could account for their inconsistency. People are voting based on how they think they should vote rather than how they feel because the survey is not anonymous. Open the Wine Quality dataset. Questions 8 - 13 concern this dataset. 8. Which variables are categorical and which are numerical? Write two questions you would try to answer about these wines using this dataset. The producer variable is categorical, and all the others are numerical. How does density affect the quality rating? What location has the overall highest quality rating? 9. Use Minitab to construct a dotplot of the Quality Rating variable in the dataset. 10. Wines are typically rated by connoisseurs on a scale of 1 – 10 (10 for best). These three wine producers all produce wines of all types, including many 9 and 10 rated wines of the highest quality. Given this, what does the dotplot in Question 9 tell you about the group of wines in this sample? (Hint: Is this a random sample of all of their wines?) It only includes wines with a quality rating ranging from 4 to 7, so it is not a complete population, just a sample. 11. Use Minitab to compute a 6-number summary of the Citric Acid variable. Compare the mean and the median, the minimum, median, and the maximum. a. What does this comparison suggest about the shape of the distribution of this variable? The data is skewed right because the median is 0.1950 while the minimum is 0.0000 and the max is 0.7000 b. Use Minitab to create a graph that will show you the shape of the distribution and give a brief description of the distribution. (Hint: Do a histogram.) 12. Construct boxplots of the Citric Acid variable grouped by Producer. (Use: Boxplot - > One Y, With Groups.) a. Which producer’s wines have the highest IQR of Citric Acid? Italy b. Which producer produced the most wines that are outliers with regard to Citric Acid? California 13. a. Standardize the Citric Acid variable using Minitab. (Use the observed (that is, the variable’s) mean and std. dev.) Store the new variable (call it: Citric Acid.-Std) in an open column of your worksheet. Write here the first 5 standardized scores. -1.22073, -1.22073, -0.99643, 1.91941, -1.22073 b. Use Minitab to show that the mean of the standardized variable is 0 and the standard deviation is 1. (Why this is true of all standardized variables when standardization is done using the observed mean and standard deviation will be explained in the Chapter 1 solutions. To see why, consider Question 1.) Variable Mea n StDe v Citric Acid.-Std -0.000 1.000 c. Use Minitab to create a histogram of the standardized variable and compare it to your histogram from Question 11b. What do you notice? Very similar but with different scales. 14. The six numbers below are drawn from N(25, 16). Compute by hand their standard scores. (Just type in your answers; remember, by-hand calculations need not be shown.) a. 2.5 b. 58.0 a.-1.40625 c. 22.5 d. 33 b.2.0625 e. 6.75 f. 48.5 c.-0.15625 d.0.5 e.-1.140625 f.1.46875 b. What relationship do you see between the number, its z-score, and the mean of the distribution? The z score is a result of taking the number and subtracting it by the mean minus the standard deviation. 15. Recall that the numbers ±1.96 cut off the center 95% of N(0, 1). (See the graph below.) Distribution Plot Normal, Mean=0, StDev=1 0.9500 0.4 Density 0.3 0.2 0.1 0.0 -1.96 0 1.96 X Recall that we say, therefore, that 1.96 is the 95% confidence multiplier for the standard normal. a. Use Minitab to create a graph showing that 1.282 is the 80% confidence multiplier for the standard normal. (Hint: Use a Probability Distribution Plot.) b. Use Minitab to find the 90% confidence multiplier for the standard normal. c. Use Minitab to create a graph showing the 80% confidence multiplier for the t with 50 degrees of freedom. Briefly explain, using what we have discussed about the t distributions, why this number is greater than the number found in part a. T distribution has heavier tails, so the T distribution is greater than the standard normal. d. Will the 90% confidence multiplier for the t with 50 degrees of freedom be greater than or less than 1.645? (Hint: You may check this with Minitab.) Explain your answer using what we have discussed about the t distributions. It will be greater than 1.299 because T distribution has heavier tails. 16. Notice that in the standard normal, the number 1.96 cuts off an upper tail of 0.025. (And, by symmetry and centering, -1.96 cuts off a lower tail of 0.025.) Recall that we say that 1.96 is the 1-sided critical value for 0.025 in the standard normal (and the 2-sided critical value for 0.05 – the sum of the upper and lower tails). (See the graph below.) Distribution Plot Normal, Mean=0, StDev=1 0.4 Density 0.3 0.2 0.1 0.025 0.0 0 1.960 X a. Use Minitab to create graphs showing the 1-sided critical values for 0.01 and 0.05 in the standard normal. Use: Graph – Probability Distribution Plot – View Probability – N(0, 1) – Shaded Area – Probability – Right Tail – (0.01 then repeat for 0.05). b. Considering the graphs you created for part a, give the 2-sided critical values for 0.02 and for 0.10 in the standard normal. Briefly explain your answers. Critical Values for 0.02 = +/- 2.326 Critical Values for 0.10 = +/- 1.645 c. Use Minitab to create graphs showing the 1-sided critical values for 0.01 and 0.05 in the t with 50 degrees of freedom. Briefly explain, using what we have discussed about the t distributions, why these values are greater than the corresponding values from part a. (Same as part a but change to t – Degrees of freedom: 50.) T distributions have heavier tails so the values are greater. d. Consider the probabilities, 0.001 and 0.005. In the standard normal, will the 1-sided critical value for 0.001 be greater than or less than the 1-sided critical value for 0.005? Briefly explain your answer. The critical value for 0.005 will be less than 0.001becasue when confidence levels decrease then critical values increase. e. Consider the probabilities, 0.001 and 0.005. In the t with 50 degrees of freedom, will the 1-sided critical value for 0.001 be greater than or less than the 1-sided critical value for 0.005? Briefly explain your answer. The critical value for 0.005 will be less than 0.001becasue when confidence levels decrease then critical values increase. 17. Suppose that, for some hypothesis test (which will discuss beginning in Chapter 5), we compute a test statistic of 1.88. Consider the graph below. It shows that the number 1.88 (picked at random for demonstration purposes) cuts off an upper tail of 0.03005 in the standard normal. Distribution Plot Normal, Mean=0, StDev=1 0.4 Density 0.3 0.2 0.1 0.03005 0.0 0 1.88 X We say that 1.88 gives a 1-sided p-value of 0.03005 in the standard normal. That is, this upper tail area (shaded in red) is called the p-value. (By symmetry and centering, the 2-sided p-value would be 0.0601 – 2 times the 1-sided p-value.) a. Using Minitab, create a graph showing the 1-sided p-value for 1.718 in the standard normal. What is the 2-sided p-value for 1.718 in the standard normal? The 2-sided p-value for 1.718 would be 0.0858 or 2 times 0.0429. b. Using Minitab, create a graph showing the 1-sided p-value for 1.718 in the t with 50 degrees of freedom. What is the 2-sided p-value for 1.718 in the t with 50 degrees of freedom? Briefly explain, using what we have discussed about the t distributions, why these p-values are larger than the corresponding ones found in part a. 2-sided p-value is 0.04599 multiplied by 2 to get 0.09198 c. In the standard normal, will the number 1.850 give a greater or lesser 1-sided pvalue than that given by 1.718? Briefly explain your answer. 1.850 will give a smaller p-value than 1.718 because it lies to the right of it on the x-axis, meaning it will have a smaller shaded area. d. In the t with 50 degrees of freedom, will the number 1.850 give a greater or lesser 1-sided p-value than that given by 1.718? Briefly explain your answer. Just like in part c, 1.850 will give a smaller p-value than 1.718 because it falls to the right of it on the x-axis. 18. Consider the Probability Distribution Plots below. The top plot is the standard normal, N(0, 1). The bottom plot is N(100, 15), the IQ distribution. a. In the top plot, which of the numbers is the test statistic? 1.696 b. In the top plot, which of the numbers is the p-value? 0.04494 c. In the bottom plot, the shaded red area (p-value) is identical to that of the top plot. Briefly explain why this is so? If you find the z-score of the bottom graph, it is 1.696. This is the test statistic of the top graph. IMPORTANT NOTE on critical values vs. p-values: - Note, a critical value is a number. It is obtained from a given probability and a distribution. It is the number that cuts the given probability in the tail (or tails, if 2-sided) of the given distribution. - A p-value is a probability. It is obtained from a number (such as a z-score or tscore, which are usually called test statistics) and is the tail probability (either 1sided or 2-sided) that is cut off by that number.