The Consumer Reports Restaurant Customer Satisfaction Survey is based upon 148,5

Question

The Consumer Reports Restaurant Customer Satisfaction Survey is based upon 148,599 visits to full-service restaurant chains (Consumer Reports website, February 11, 2009). Assume the following data are representative of the results reported. The variable type indicates whether the restaurant is an Italian restaurant or a seafood/steakhouse. Price indicates the average amount paid per person for dinner and drinks, minus the tip. Score reflects diners' overall satisfaction, with higher values indicating greater overall satisfac- tion. A score of 80 can be interpreted as very satisfied. Type Italian Seafood/Steakhouse Seafood/Steakhouse Restaurant Price () 16 24 26 Score Bertucci's Black Angus Steakhouse Bonefish Grill 79 85 continued)

Explanation / Answer

> score <- c(77,79,85,84,81,77,86,75,83,71,81,76,81,83,81,81,80,78,82,79,76)

> price <- c(16,24,26,18,17,18,23,17,28,15,17,17,19,22,16,19,20,18,18,12,16)

(a)

> lm(score ~ price)

Call:
lm(formula = score ~ price)

Coefficients:
(Intercept) price  
69.2760 0.5586

satisfaction score = 69.2760 + 0.5586 * price

(b)

> summary(lm(score ~ price))

Call:

lm(formula = score ~ price)

Residuals:

Min 1Q Median 3Q Max

-6.655 -2.213 1.111 2.228 4.669

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 69.2760 3.4005 20.373 2.28e-14

price 0.5586 0.1769 3.158 0.00518

(Intercept) ***

price **

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Signif. codes:  

0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 3.026 on 19 degrees of freedom

Multiple R-squared: 0.3442, Adjusted R-squared: 0.3097

F-statistic: 9.971 on 1 and 19 DF, p-value: 0.005182

The p-value to test the null hypothesis that there is no relationship between overall satisfaction and average meal price = 0.005182

Level of significance = 0.05

which is higher than p-value

So,we reject the null hypothesis at 0.05 level of significance

Hence, we conclude that there is a significant relationship between overall satisfaction and average meal price

(c)

> type <- c(1,0,0,1,1,0,1,0,1,0,1,0,0,1,0,1,0,0,1,1,1)

type is a dummy variable for the type of the restaurant -> 1 = italian, 0 = seafood/steakhouse

(d)

> lm(score ~ price + type)

Call:
lm(formula = score ~ price + type)

Coefficients:
(Intercept) price type  
67.4049 0.5734 3.0382  

score = 67.4049 + 0.5734 * price + 3.0382 * Indicator (type = Italian)

(e)

> summary(lm(score ~ price + type))

Call:

lm(formula = score ~ price + type)

Residuals:

Min 1Q Median 3Q Max

-5.0063 -2.1531 0.2734 1.6758 4.4203

Coefficients:

Estimate Std. Error t value Pr(>|t|)   

(Intercept) 67.4049 3.0535 22.07 1.74e-14 ***

price 0.5734 0.1546 3.71 0.0016 **

type 3.0382 1.1552 2.63 0.0170 *  

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Signif. codes:  

0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 2.642 on 18 degrees of freedom

Multiple R-squared: 0.5262, Adjusted R-squared: 0.4736

F-statistic: 9.997 on 2 and 18 DF, p-value: 0.001202

The p-value to test the significance of factor - type of restaurant in overall customer satisfaction = 0.0170

which is less than 0.05 and is staistically significant

hence, the type of restaurant is a significant factor in overall customer satisfaction

(f)

price = 20

type = 0 (since type = seafood/steakhouse)

estimated customer satisfaction score

= 67.4049 + 0.5734 * price + 3.0382 * Indicator (type = Italian)

= 67.4049 + 0.5734 * 20 + 3.0382 * 0

= 78.8729

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