Qnt561_Wk_6_Individual_Assignment

Exercise 88 Refer to the Baseball 2005 data, which reports information on the 30 major league teams for the 2005 baseball season. a. Select the variable team salary and find the mean, median, and the standard deviation. Mean = 73.06 Median = 66.20 Standard deviation = 34.23 b. Select the variable that refers to the age the stadium was built. (Hint: Subtract the year in which the stadium was built from the current year to find the stadium age and work with that variable.) Find the mean, median, and the standard deviation. Mean = 28.20 Median = 17.50 Standard deviation = 25.94 c. Select the variable that refers to the seating capacity of the stadium. Find the mean, median, and the standard deviation. Mean = 45,913 Median = 44,174 Standard deviation = 5,894 Exercise 56 Assume the likelihood that any flight on Northwest Airlines arrives within 15 minutes of the scheduled time is .90. We select four flights from yesterday for study. a. What is the likelihood all four of the selected flights arrived within 15 minutes of the scheduled time' P(x=4) = 0.65610 b. What is the likelihood that none of the selected flights arrived within 15 minutes of the scheduled time' P(x=0) = 0.00010 c. What is the likelihood at least one of the selected flights did not arrive within 15 minutes of the scheduled time' P(x≥1) = 0.99990 n = 4, p = 0.9, q = 0.1 Binomial Distribution x p(x) 0 0.00010 1 0.00360 2 0.04860 3 0.29160 4 0.65610 Exercise 64 An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of two emails per hour. Assume the arrival of these emails are approximated by the Poisson distribution. a. What is the probability Linda Lahey, company president, received exactly 1 email between 4 P.M. and 5 P.M. yesterday' P(x=1) = 0.270670566 b. What is the probability she received 5 or more email during the same period' P(x≥5) = 0.016563608 c. What is the probability she did not receive any email during the period' P(x=0) = 0.135335283 Exercise 50 Fast Service Truck Lines uses the Ford Super Duty F-750 exclusively. Management made a study of the maintenance costs and determined the number of miles traveled during the year followed the normal distribution. The mean of the distribution was 60,000 miles and the standard deviation 2,000 miles. Mean = 60,000, St dev = 2000, Normal Distribution a. What percent of the Ford Super Duty F-750s logged 65,200 miles or more' P(x≥65200) = 0.004661188 = 0.4661% b. What percent of the trucks logged more than 57,060 but less than 58,280 miles' P(57,060 25) = 0.016947427 b. What is the likelihood the sample mean is greater than $22.50 but less than $25.00' P(22.5 < x-bar < 25) = 0.90440297 c. Within what limits will 90 percent of the sample means occur' Left Limit = 22.33691285 = $22.34 Right Limit = 24.66308715 = $24.66 Exercise 54 Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 20 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an employer averaged $10,979. The standard deviation of the sample was $1,000. n=20; x-bar = 10,979, s = 1000; degrees of freedom (df) = 19 a. Based on this sample information, develop a 90 percent confidence interval for the population mean yearly premium. Confidence = .90, so α = .10 and α/2 = .05 t(df=19; α/2 = .05) = 1.729132792 Error = (1.729)(1000)/√20 = 386.6161533 = n = 107 Interval = x-bar ± Error Interval is (10592.38, 11365.62) b. How large a sample is needed to find the population mean within $250 at 99 percent confidence' Confidence = .99; Error = 250; α=.01 so α/2 = .005 Z(.005) = 2.575829304 n = [(2.579)(1000)/250]^2 = 106.1583456 Exercise 42 During recent seasons, Major League Baseball has been criticized for the length of the games. A report indicated that the average game lasts 3 hours and 30 minutes. A sample of 17 games revealed the following times to completion. (Note that the minutes have been changed to fractions of hours, so that a game that lasted 2 hours and 24 minutes is reported at 2.40 hours.) 2.98 2.40 2.70 2.25 3.23 3.17 2.93 3.18 2.80 2.38 3.75 3.20 3.27 2.52 2.58 4.45 2.45 Ho: µ ≥ 3.5 H1: µ < 3.5 α = .05 p-value = 0.001 At the .05 level of significance, we reject Ho because p-value (.001) < α (.05). Thus, there is enough sample evidence to support the claim that the average baseball game lasts less than 3.5 hours. Exercise 58 The amount of income spent on housing is an important component of the cost of living. The total costs of housing for homeowners might include mortgage payments, property taxes, and utility costs (water, heat, electricity). An economist selected a sample of 20 homeowners in New England and then calculated these total housing costs as a percent of monthly income, five years ago and now. The information is reported below. Is it reasonable to conclude the percent is less now than five years ago' (µ1 represents now and µ2 represents 5 years ago) Ho: µ1 ≥ µ2 H1: µ1 < µ2 α = .05 p-value = 0.036 At the .05 level of significance, we reject Ho because p-value (.036) < α (.05). Thus, there is enough sample evidence to support the claim that the percent now is less than the percent five years ago. Exercise 42 (Ch. 12) Martin Motors has in stock three cars of the same make and model. The president would like to compare the gas consumption of the three cars (labeled car A, car B, and car C) using four different types of gasoline. For each trial, a gallon of gasoline was added to an empty tank, and the car was driven until it ran out of gas. The following table shows the number of miles driven in each trial. Using the .05 level of significance: a. Is there a difference among types of gasoline' (µ1 = Car A; µ2 = Car B; µ3 = Car C) Ho: µ1 = µ2 = µ3 H1: At least one mean differs from another α = .05 p-value = 0.424 At the .05 level of significance, we fail to reject Ho because p-value (.424) > α (.05). Thus, there is not enough sample evidence to show that there is a difference among the cars. b. Is there a difference in the cars' (µ1 = Regular; µ2 = Unleaded; µ3 = Super; µ4 = Premium) Ho: µ1 = µ2 = µ3 = µ4 H1: At least one mean differs from another α = .05 p-value = 0.166 At the .05 level of significance, we fail to reject Ho because p-value (.166) > α (.05). Thus, there is not enough sample evidence to show that there is a difference among types of gasoline. Exercise 37 A regional commuter airline selected a random sample of 25 flights and found that the correlation between the number of passengers and the total weight, in pounds, of luggage stored in the luggage compartment is 0.94. Using the .05 significance level, can we conclude that there is a positive association between the two variables' n=25 r = 0.94; α=.05 Ho: ρ = 0 H1: ρ ≠ 0 t = r √((n-2)/(1-r^2)) = 13.21342127 p-value = 0 At the .05 level of significance, we reject Ho because p-value (0) < α (.05). Thus, there is enough sample evidence to show that there is a relationship between the two variables. Exercise 40 A suburban hotel derives its gross income from its hotel and restaurant operations. The owners are interested in the relationship between the number of rooms occupied on a nightly basis and the revenue per day in the restaurant. Below is a sample of 25 days (Monday through Thursday) from last year showing the restaurant income and number of rooms occupied. Use a statistical software package to answer the following questions. a. Does the breakfast revenue seem to increase as the number of occupied rooms increases' Draw a scatter diagram to support your conclusion. Yes. There mostly appears to be an increase in the breakfast revenue as the number of rooms occupied increases. The scatter plot supports this claim. b. Determine the coefficient of correlation between the two variables. Interpret the value. r = 0.393, which appears to be a small positive correlation c. Is it reasonable to conclude that there is a positive relationship between revenue and occupied rooms' Use the .10 significance level. At the .10 level of significance, we reject Ho because p-value (.052) < α (.10). Thus, there is enough sample evidence to show that there is a positive relationship between breakfast revenue and rooms occupied. d. What percent of the variation in revenue in the restaurant is accounted for by the number of rooms occupied' 15.5% of the breakfast revenue variation is accounted for by the number of rooms occupied in the hotel. n=25; α=.10 Ho: ρ = 0 H1: ρ ≠ 0 Regression Analysis: income versus occupied The regression equation is income = 1387 + 1.37 occupied Predictor Coef SE Coef T P Constant 1386.56 25.19 55.04 0.000 occupied 1.3671 0.6662 2.05 0.052 S = 40.1912 R-Sq = 15.5% R-Sq(adj) = 11.8% Analysis of Variance Source DF SS MS F P Regression 1 6801 6801 4.21 0.052 Residual Error 23 37153 1615 Total 24 43954 Unusual Observations Obs occupied income Fit SE Fit Residual St Resid 2 47.0 1361.00 1450.82 10.95 -89.82 -2.32R 13 54.0 1537.00 1460.39 14.53 76.61 2.04R R denotes an observation with a large standardized residual. Correlations: income, occupied Pearson correlation of income and occupied = 0.393 P-Value = 0.052 Exercise 17 The district manager of Jasons, a large discount electronics chain, is investigating why certain stores in her region are performing better than others. She believes that three factors are related to total sales: the number of competitors in the region, the population in the surrounding area, and the amount spent on advertising. From her district, consisting of several hundred stores, she selects a random sample of 30 stores. For each store she gathered the following information. Y = total sales last year (in $ thousands). X1 = number of competitors in the region. X2 = population of the region (in millions). X3 = advertising expense (in $ thousands). The sample data were run on MINITAB, with the following results. Analysis of variance SOURCE DF SS MS Regression 3 3050.00 1016.67 Error 26 2200.00 84.62 Total 29 5250.00 Predictor Coef StDev t-ratio Constant 14.00 7.00 2.00 X1 _1.00 0.70 _1.43 X2 30.00 5.20 5.77 X3 0.20 0.08 2.50 Regression equation: y = 14 - x1 + 30x2 + .20x3 a. What are the estimated sales for the Bryne store, which has four competitors, a regional population of 0.4 (400,000), and advertising expense of 30 ($30,000)' a. y = 14 - 4 + 30(0.4) + .20(30) = 28 b. Compute the value. r^2 = (Sstotal - Sserror)/Sstotal = 0.580952381 c. Compute the multiple standard error of estimate. Ho: ρ = 0 H1: ρ ≠ 0 F = Msreg/Mserror = 12.01453557 p-value = 0.0000402 At the .05 level of significance, we reject Ho because p-value (0) < α (.05). Thus, there is enough sample evidence to show that the regression coefficients are not equal to zero. Ho: ρ1 = 0 H1: ρ1 ≠ 0 Ho: ρ2 = 0 H1: ρ2 ≠ 0 Ho: ρ3 = 0 H1: ρ3 ≠ 0 pvalue1 = 0.288979466 pvalue2 = 0.028747521 pvalue3 = 0.129611728 d. Conduct a global test of hypothesis to determine whether any of the regression coefficients are not equal to zero. Use the .05 level of significance. At the .05 level of significance, the population coefficient is significant. However, the competitor and advertising coefficients are not significant at the .05 level of significance. Therefore, we should consider eliminating these two insignificant variables. e. Conduct tests of hypotheses to determine which of the independent variables have significant regression coefficients. Which variables would you consider eliminating' Use the .05 significance level. At the .05 level of significance, the regression variables are significant, therefore we should consider eliminating these two insignificant variables. Exercise 18 Suppose that the sales manager of a large automotive parts distributor wants to estimate as early as April the total annual sales of a region. On the basis of regional sales, the total sales for the company can also be estimated. If, based on past experience, it is found that the April estimates of annual sales are reasonably accurate, then in future years the April forecast could be used to revise production schedules and maintain the correct inventory at the retail outlets. Several factors appear to be related to sales, including the number of retail outlets in the region stocking the company’s parts, the number of automobiles in the region registered as of April 1, and the total personal income for the first quarter of the year. Five independent variables were finally selected as being the most important (according to the sales manager). Then the data were gathered for a recent year. The total annual sales for that year for each region were also recorded. Note in the following table that for region 1 there were 1,739 retail outlets stocking the company’s automotive parts, there were 9,270,000 registered automobiles in the region as of April 1 and so on. The sales for that year were $37,702,000. a. Consider the following correlation matrix. Which single variable has the strongest correlation with the dependent variable' The correlations between the independent variables outlets and income and between cars and outlets are fairly strong. Could this be a problem' What is this condition called' The independent variable Income has the highest correlation with the dependent variable sales, with a correlation of 0.964. When independent variables have a strong correlation with each other (as opposed to the dependent variable), it is called multicollinearity. This can be a problem because multicollinearity causes the coefficients for these independent predictors to change dramatically with small changes in the data. Thus, it makes their coefficients unreliable. b. The output for all five variables is on the following page. What percent of the variation is explained by the regression equation' R^2 = .994, so 99% of the variance is accounted for by the regression equation. c. Conduct a global test of hypothesis to determine whether any of the regression coefficients are not zero. Use the .05 significance level. Ho: β1 = β2 = β3 = β4 = β5 = 0 H1: β1 ≠ β2 ≠ β3 ≠ β4 ≠ β5 ≠ 0 F = MSreg/Mserr = 140.4229075 p-value = 0.0000000294 At the .05 level of significance, we reject Ho because p-value (0) < α (.05). Thus, there is enough sample evidence to show that the regression coefficients are not equal to zero. d. Conduct a test of hypothesis on each of the independent variables. Would you consider eliminating “outlets” and “bosses”' Use the .05 significance level. Ho: β1 = 0 H1: β1 ≠ 0 Ho: β2 = 0 H1: β2 ≠ 0 Ho: β3 = 0 H1: β3 ≠ 0 Ho: β4 = 0 Ho: β5 = 0 H1: β5 ≠ 0 H1: β4 ≠ 0 pvalue1 = 0.814902011 pvalue2 = 0.03451402 pvalue3 = 0.000722628 pvalue4 = 0.082939037 pvalue5 = 0.873261877 At the .05 level of significance, the cars, income and age coefficients are significant. However, the outlets and bosses coefficients are not significant at the .05 level of significance. Thus, we should consider eliminating these two insignificant variables. e. The regression has been rerun below with “outlets” and “bosses” eliminated. Compute the coefficient of determination. How much has changed R2 from the previous analysis' The R^2 did not change when the two variables were omitted from the regression. It is still 0.994, which is 99% of the variance explained by the regression equation. f. Following is a histogram and a stem-and-leaf chart of the residuals. Does the normality assumption appear reasonable' Yes, the residuals appear to be normally distributed, so the normality assumption appears very reasonable. g. Following is a plot of the fitted values of Y (i.e., and the residuals. Do you see any violations of the assumptions' The residuals about the fitted values appear to be scattered and random, meaning it does not appear that any assumptions have been violated. Exercise 22 Banner Mattress and Furniture Company wishes to study the number of credit applications received per day for the last 300 days. The information is reported on the next page. To interpret, there were 50 days on which no credit applications were received, 77 days on which only one application was received, and so on. Would it be reasonable to conclude that the population distribution is Poisson with a mean of 2.0' Use the .05 significance level. Hint: To find the expected frequencies use the Poisson distribution with a mean of 2.0. Find the probability of exactly one success given a Poisson distribution with a mean of 2.0. Multiply this probability by 300 to find the expected frequency for the number of days in which there was exactly one application. Determine the expected frequency for the other days in a similar manner. Ho: the distribution is Poisson H1: the distribution is not the Poisson distribution # of apps Obs Freq Poisson Prob Expected (O-E)^2/E 0 50 0.135335283 40.60058497 2.176052 1 77 0.270670566 81.20116994 0.217359 2 81 0.270670566 81.20116994 0.000498 3 48 0.180447044 54.13411329 0.695076 4 31 0.090223522 27.06705665 0.571471 5+ 13 0.052653017 15.7959052 0.494881 X2 = 4.155338 p-value = 0.472725119 At the .05 level of significance, we fail to reject Ho because p-value (.4728) > α (.05). Thus, there is sufficient sample evidence to support that the claim that the distribution is a Poisson distribution with a mean of 2.0.