Question

theres 4 total questions

oare Was teR Mailed to these tive cardhoiders requesting intormation on the numbe time period. The data follow. Sick Days & Age iles 1,000's)K$1,000's) harges Cardholder 1.2 2.0 1.5 2.3 2.7 1.6 2.5 2.2 4.2 3.2 2.7 0.5 2.1 0.8 reA regression Analysis has been performed to estimate the model and the output is given. Regression Statistics Multiple R R Square Adjusted R Square ,19954 56026 0.31389 71405 tandard Error bservations ANOVA O 2019 01:41:34 A regression Analysis has been performed to estimate the model and the output is given. Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations 0.56026 31389 19954 71405 ANOVA nificance 14864 11.39955 3.05918 7 4.45873 1,39955 .50986 74495 Regression Residual Total ev upper 95% 7204 Lower 95% 1585 73004 -value 00169 .14864 Stat 38633 1.65679 Coefficients 3.69394 Standard Error intercept Miles 310 .421 58 step 1 of 4: What is the slope of the estimated regression equation? (Report to 2 decimal places.) Keypad Tables Answer 1 Point 019 Hawkes Learning 74.45873 Total Coefficients Standard Error t Stat value 00169 14864 69394 0.69847 68580 42158 Intercept Miles 38633 1.65679 Step 1 of 4: What is the slope of the estimated regression equation? (Report to 2 decimal places.) Answer 1 Point 9 Hawkes Learning ANOVA MS ignificance 14864 Regression Residual Total 1.39955 6 3.05918 4.45873 1.39955 2.74495 50986 Standard Error 0.68580 0,42158 P-value 00169 14864 Coefficients 3.69394 0.69847 over 95% 01585 1.73004 t Stat pper 95% 5.37204 ntercept Miles 5.38633 1.65679 33310 Step 2 of 4: What fraction of the variation in credit card charges is explained by miles traveled? (Report as a decimal fraction to 2 decimal place m Tables Keypad Answer 1 Point wkes Learning ANOVA MS 1.39955 f SS ignificance F 14864 11.39955 6 3.05918 74.45873 74495 egression esidual otal 50986 Lower 95% 2.01585 P-value 00169 14864 Stat 38633 1.65679 Coefficients 3.69394 0.69847 Standard Error 0.68580 ntercept Miles 1.73004 42158 Step 3 of 4: Determine the 95% confidence interval for ß-the slope of the line. (Report to 3 decimal places.) Tables Answer Point ANOVA df ss MS 1.39955 0.50986 Significance F 0.14864 egression esidual otal 11.39955 6 3.05918 45873 2.74495 Upper 9 oefficients 69394 t Stat 5.38633 1.65679 Standard Error 0.68580 0.42158 value 00169 14864 Lower 95% 2.01585 Intercept Miles 37204 0.33310 0.69847 1.73004 Step 4 of 4: Determine the 95% confidence interval for the average credit card charges, given that the cardholder traveled 2 thousand miles. (Report answer to 2 decimal places.) Answer 1 Point Thousand Dollars Lower lawkes Learning

Answer 1

Q1: Slope of the estimated regression equation= -0.69847 = -0.70

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Q2: Fraction of the variation in credit card charges that is explained by miles traveled = 0.3139 = 0.31

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Q3: 95% confidence interval for β₁ :

Lower = -1.730

Upper = 0.333

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Q4: Regression equation :               
ŷ = 3.69394 - 0.69847 x

Predicted value at X = 2      
ŷ = 3.69394 - 0.69847 * 2 = 2.297   

Standard error, se = 0.71405

Standard error for slope, se_b = 0.42158

x̅ = Ʃ x /n = 1.5125

At α = 0.05 and df = n-2 = 6, critical value, t_c = T.INV.2T(0.05, 6) =  2.4469

95% Confidence interval for the average credit card charges:

tc * se n se/se) 1 (2 -1.5125)2 2.297 ± 2.4469 * 0.71405, + 8 (0.71405/0.42158)2 1.50, 3.09)

Lower: 1.50 Thousand Dollars.

Upper: 3.09 Thousand Dollars.