B. Would the Floor Area be a significant predictor of Assessed Value?
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- A simple random sample of size n is drawn. The sample mean, x, is found to be 18.2, and the sample standard deviation, s, is found to be 4.5. Click the icon to view the table of areas under the t-distribution. (a) Construct a 95% confidence interval about u if the sample size, n, is 35. Lower bound: ; Upper bound: (Use ascending order. Round to two decimal places as needed.)A special bumper was installed on selected vehicles in a large fleet. The dollar cost of body repairs was recorded for all vehicles that were involved in accidents over a 1-year period. Those with the special bumper are the test group and the other vehicles are the control group, shown below. Each "repair incident" is defined as an invoice (which might include more than one separate type of damage). Statistic Test Group Control Group Mean Damage x¯1x¯1 = $ 1,101 x¯2x¯2 = $ 1,766 Sample Standard Deviation s1 = $ 696 s2 = $ 838 Repair Incidents n1 = 12 n2 = 9 (a) Construct a 90 percent confidence interval for the true difference of the means assuming equal variances. (Round answers to 3 decimal places. Negative values should be indicated by a minus sign.) (b) Repeat part (a), using the assumption of unequal variances with Welch's formula for d.f. (Round answers to 2 decimal places. Negative values should be indicated by a minus sign.) (d) Construct separate…A special bumper was installed on selected vehicles in a large fleet. The dollar cost of body repairs was recorded for all vehicles that were involved in accidents over a 1-year period. Those with the special bumper are the test group and the other vehicles are the control group, shown below. Each "repair incident" is defined as an invoice (which might include more than one separate type of damage). Statistic Test Group Control Group Mean Damage X¯¯¯1X¯1 = $ 1,153 X¯¯¯2X¯2 = $ 1,751 Sample Std. Dev. s1 = $ 663 s2 = $ 820 Repair Incidents n1 = 16 n2 = 14 Source: Unpublished study by Thomas W. Lauer and Floyd G. Willoughby. (a) Construct a 99 percent confidence interval for the true difference of the means assuming equal variances. (Round your final answers to 3 decimal places. Negative values should be indicated by a minus sign.) The 99% confidence interval is from to (b) Repeat part (a), using the assumption of unequal variances with Welch's…
- Conduct a global test on the set of independent variables. Interpret Regression Statistics Multiple R 0.87027387 R Square 0.75737661 Adjusted R Square 0.75615535 Standard Error 14.6932431 Observations 600 ANOVA df SS MS F Significance F Regression 3 401662.063 133887.354 620.160683 8.708E-183 Residual 596 128671.271 215.891394 Total 599 530333.333 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept -3.9995369 3.05935528 -1.3073137 0.19161031 -10.007965 2.00889078 -10.007965 2.00889078 Annual Income 0.0002132 3.1402E-05 6.78944156 2.7269E-11 0.00015153 0.00027487 0.00015153 0.00027487 Married 45.7808695 1.20203164 38.0862434 8.444E-162 43.4201368 48.1416023 43.4201368 48.1416023 Male 21.9175699 1.20122625 18.2459964 2.0045E-59…A special bumper was installed on selected vehicles in a large fleet. The dollar cost of body repairs was recorded for all vehicles that were involved in accidents over a 1-year period. Those with the special bumper are the test group and the other vehicles are the control group, shown below. Each "repair incident" is defined as an invoice (which might include more than one separate type of damage). Statistic Test Group Control Group Mean Damage X¯¯¯1X¯1 = $ 1,077 X¯¯¯2X¯2 = $ 1,800 Sample Std. Dev. s1 = $ 693 s2 = $ 814 Repair Incidents n1 = 16 n2 = 13 Source: Unpublished study by Thomas W. Lauer and Floyd G. Willoughby. (a) Construct a 90 percent confidence interval for the true difference of the means assuming equal variances. (Round your final answers to 3 decimal places. Negative values should be indicated by a minus sign.) The 90% confidence interval is from to (b) Repeat part (a), using the assumption of unequal variances with Welch's…A special bumper was installed on selected vehicles in a large fleet. The dollar cost of body repairs was recorded for all vehicles that were involved in accidents over a 1-year period. Those with the special bumper are the test group and the other vehicles are the control group, shown below. Each "repair incident" is defined as an invoice (which might include more than one separate type of damage). Statistic Test Group Control Group Mean Damage X⎯⎯⎯1X¯1 = $ 1,245 X⎯⎯⎯2X¯2 = $ 1,790 Sample Std. Dev. s1 = $ 721 s2 = $ 835 Repair Incidents n1 = 17 n2 = 14 Source: Unpublished study by Thomas W. Lauer and Floyd G. Willoughby. (a) Construct a 99 percent confidence interval for the true difference of the means assuming equal variances. (Round your final answers to 3 decimal places. Negative values should be indicated by a minus sign.) The 99% confidence interval is from to (b) Repeat part (a), using the assumption of unequal variances with Welch's…
- A special bumper was installed on selected vehicles in a large fleet. The dollar cost of body repairs was recorded for all vehicles that were involved in accidents over a 1-year period. Those with the special bumper are the test group and the other vehicles are the control group, shown below. Each "repair incident" is defined as an invoice (which might include more than one separate type of damage). Statistic Test Group Control Group Mean Damage X⎯⎯⎯1X¯1 = $ 1,188 X⎯⎯⎯2X¯2 = $ 1,792 Sample Std. Dev. s1 = $ 712 s2 = $ 800 Repair Incidents n1 = 16 n2 = 14 Source: Unpublished study by Thomas W. Lauer and Floyd G. Willoughby. (a) Construct a 95 percent confidence interval for the true difference of the means assuming equal variances. (Round your final answers to 3 decimal places. Negative values should be indicated by a minus sign.) The 95% confidence interval is from to (b) Repeat part (a), using the assumption of unequal variances with Welch's…Predicted index score.Iron-deficiency anemia is an important nutritional healthproblem in the United States. A dietary assessment wasperformed on 51 boys 9 to 11 years of age whose familieswere below the poverty level. The mean daily iron intakeamong these boys was found to be 12.50 mg with standarddeviation 4.75 mg. Suppose the mean daily iron intakeamong a large population of 9- to 11-year-old boys fromall income strata is 14.44 mg. We want to test whether themean iron intake among the low-income group is differentfrom that of the general population. The standard deviation of daily iron intake in the larger population of 9- to 11-year-old boys was 5.56 mg. We want totest whether the standard deviation from the low-incomegroup is comparable to that of the general population.7.38) State the hypotheses that we can use to answer thisquestion.7.39) Carry out the test in Problem 7.38 using the criticalvalue method with an α level of .05, and summarize yourfindings.7.40) What is the p-value for the test…
- Iron-deficiency anemia is an important nutritional healthproblem in the United States. A dietary assessment wasperformed on 51 boys 9 to 11 years of age whose familieswere below the poverty level. The mean daily iron intakeamong these boys was found to be 12.50 mg with standarddeviation 4.75 mg. Suppose the mean daily iron intakeamong a large population of 9- to 11-year-old boys fromall income strata is 14.44 mg. We want to test whether themean iron intake among the low-income group is differentfrom that of the general population. The standard deviation of daily iron intake in the larger pop-ulation of 9- to 11-year-old boys was 5.56 mg. We want to test whether the standard deviation from the low-income group is comparable to that of the general population.Problem a) State the hypotheses that we can use to answer thisquestion. Problem b) Carry out the test in Problem a using the critical-value method with an α level of 0.05, and summarize your findings. Problem c) What is the p-value…indicator for gender (=1 if male, =0 if female). Your data set has 200 observations. donation = 1.57 + 0.12age + 0.08age? – 0.43 gender + 0.07(gender * age), R? = 0.33 (0.15)* (0.67) (0.11) (0.01) (0.06) donation = 1.59 (0.68) 0.46 gender, R² = 0.30 (0.17) 1. Assuming homoscedastic errors, test the hypothesis that age affects charitable donations at the 5% level. 2. Interpret the coefficient on gender interacted with age. 3. How would you test that the difference in giving between men and women is independent of age? If you can run the test, run it. If you cannot, explain what you need.Calculate the residual for the first observation in the dataset.