Chapter 18.3, Problem 18.14E

Calculate Karl Pearson's coeffcient of skewness from the following data:Weekly Sales (Ksh) 10-12 12-14 14-16 16-18 18-20 20-22 22-24 24-26Number of Companies 12 18 35 42 50 30 8 4

A researcher hypothesizes that in a certain country the net annual growth of private sector purchases of government bonds, B, is positively related to the nominal rate of interest on the bonds, NI, and negatively related to the rate of inflation Π: Bt = a0 + a1NIt + a2Π t + ut Note that it may be hypothesized that B depends on the real rate of interest on bonds, R, where R = NI – Π.  Using a sample of 56 annual observations, s/he estimates the following equations:   (1)        Bt = 0.43 + 0.90NIt - 0.97Πt      R21 = 0.962, SSR1 = 2.20, QRESET(F1,52) = 16.6                                (3.58)   (8.80)    (-1.05)    (2)        Bt = 0.44 + 0.94Rt                   R22 = 0.960, SSR2 = 2.22, QRESET(F1,53) = 0.9                                (9.70)   (16.7)        (3)        Bt = 0.44 + 1.14NIt                  SSR3 = 9.20, QRESET(F1,53) = 59.9                                (8.84)   (36.1)    (4)        NIt = 0.08 + 0.94Πt                  R24 = 0.997, SSR4 = 0.18,  QRESET(F1,53) = 1.4…

The following table shows the annual expenditures, in dollars, per customer unit for residential landline phone services and cellular phone services in the United States in the given year.† Year Landline Cell 2004 592 378 2006 542 524 2008 467 643 2010 401 760 Calculate the regression line for each type of service. (Let t be the time in years since 2004, L be the operating revenue of landline phone services and C be the expenditure of cellular services. Round your regression parameters to two decimal places.) L(t) =        C(t) =        Determine the expenditure level at which the two lines cross. Round your answer for the expenditure level to one decimal place. million dollars

A number of studies have shown lichens (certain plants composed of an alga and a fungus) to be excellent bioindicators of air pollution. The article “The Epiphytic Lichen Hypogymnia physodes as a Biomonitor of Atmospheric Nitrogen and Sulphur Deposition in Norway” (Environ. Monitoring Assessment, 1993: 27–47) gives the following data (read from a graph) on x ¼ NO3 wet deposition (g N/m2 ) and y ¼ lichen N (% dry weight): (refer to chart)  The author used simple linear regression to analyze the data. Use the accompanying MINITAB output to answer the following questions: a. What are the least squares estimates of b0 and b1? b. Predict lichen N for an NO3 deposition value of .5. c. What is the estimate of s? d. What is the value of total variation, and how much of it can be explained by the model relationship?

Consider the following data relating hours spent studying (X) and average grade on course quizzes (Y):                                                                 X                             Y                                                                 5                              6                                                                 3                              8                                                                 4                              8                                                                 7                              10                                                                 5                              7                                                                 6                              9   Compute:    ExEy/n     1140     6     240     235

The data given below indicate the existence of a linear relationship between the x and y variables. Suppose an analyst who prepared the solutions and carried out the RI measurements was not skilled and as a result of poor technique, allowed intermediate errors to appear. The results are the following:Concentration of solution in percent (x) 10 26 33 50 61Refractive indices (y) 1.497 1.493 1.485 1.478 1.477Step 1. Carefully plot the given x and y values (from the table) on a regular graphing paper. Label then connect the points to observe a zigzag plot due to the scattered points. Step 2: Copy and fill the table given below: x (x - x̄) (x - x̄) 2 y (y - ȳ) (y - ȳ) 2 (x - x̄) (y - ȳ) 10 1.497  26 1.49333 1.48550 1.47861 1.477∑ =  ∑ = ∑ = ∑ =  ∑ = ∑ = ∑ =x̄= ∑xi ÷ Nx̄=  ȳ = ∑yi ÷ Nȳ =  Step 3. After completing the table, present following computations and the interpretation.a. Calculate the correlation coefficient (r), using the working formula: r =Σ (x − x ) (y − ȳ)√(Σ(x − x )2)(Σ(y −…

The amount of time a student devotes to class attendance and revision: and the grade obtained are assumed to be linearly related. In a small class of 15 students, the following results were obtained for this linear relationship. Y = 30 +2.50.Y (0.55) (2.96) R'= 0.85 The standard errors are in parenthesis.  a,Test the hypothesis that the intereept and slope are individually equal to zero (0) at the S% level of signiticance.                  (b) Construct n 95% conlidence interval for the true slope.                              (c)Compute the elasticity of grades with respect to time input and interpret your results.                                                        (d) Interpret your R' and explain how it relates to the slope of the regression line.  (e) From the results and your answers in (a) to (d), does time input into class attendance and revision really influence the final grade?

A major credit card company is interested in whether there is a linear relationship between its internal rating of a customer’s credit risk and that of an independent rating agency. The company collected a random sample of 200 customers and used the data to test the claim that there is a linear relationship. The following hypotheses were used to test the claim. H0:β1=0Ha:β1≠0 The test yielded a t-value of 3.34 with a corresponding p-value of 0.001. Which of the following is the correct interpretation of the p-value? If the alternative hypothesis is true, the probability of observing a test statistic at least as extreme as 3.34 is 0.001. If the alternative hypothesis is true, the probability of observing a test statistic at least as extreme as 3.34 is 0.001. A If the alternative hypothesis is true, the probability of observing a test statistic of 3.34 or greater is 0.001. If the alternative hypothesis is true, the probability of observing a test statistic of 3.34 or greater…

Based on the below table, compute the regression line that predicts Y from X. (relevant section) MX MY sX sY r 10 12 2.5 3.0 -0.6