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A random sample of 502 Vail Resorts' guests were asked to rate their satisfaction on various attributes of their visit on a scale of 1-5 with 1= very unsatisfied and 5= very satisfied. The regression model was Y= overall satisfaction score, X1X_1= lift line wait, X2X_2= amount of ski trail grooming, X3X_3= ski patrol visibility, and X4X_4= friendliness of guest services. (a) Calculate the t statistic for each coefficient to test for βj=0\beta_j=0. (b) Look up the critical value of Student's t in Appendix D for a two-tailed test at α=.01\alpha=.01. Which coefficients differ significantly from zero? (c) Use Excel to find a p-value for each coefficient.

 Predictor  Coefficient  SE  Intercept 2.89310.3680 LiftWait 0.15420.0440 AmountGroomed 0.24950.0529 SkiPatroMisibility 0.05390.0443 FriendlinessHosts 0.11960.0623\begin{array}{lcc} \hline \text { Predictor } & \text { Coefficient } & \text { SE } \\ \hline \text { Intercept } & 2.8931 & 0.3680 \\ \text { LiftWait } & 0.1542 & 0.0440 \\ \text { AmountGroomed } & 0.2495 & 0.0529 \\ \text { SkiPatroMisibility } & 0.0539 & 0.0443 \\ \text { FriendlinessHosts } & -0.1196 & 0.0623 \\ \hline \end{array}

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a)

To calculate t-value for the Intercept predictor divide the Intercept Coefficient by the Intercept standard error value:

Intercept:tcalc=2.89310.36807.8617Intercept: t_{\text{calc}}=\frac{2.8931}{0.3680}\approx 7.8617

To calculate t-value for the LiftWait predictor divide the LiftWait Coefficient by the LiftWait standard error value:

LiftWait:tcalc=0.15420.04403.5045LiftWait: t_{\text{calc}}=\frac{0.1542}{0.0440}\approx 3.5045

To calculate t-value for the AmountGroomed predictor divide the AmountGroomed Coefficient by the AmountGroomed standard error value:

AmountGroomed:tcalc=0.24950.05294.7164AmountGroomed:t_{\text{calc}}=\frac{0.2495}{0.0529}\approx 4.7164

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