Question

In: Statistics and Probability

Develop a simple linear regression model to predict the Cost of Living Index based upon Restaurant...

Develop a simple linear regression model to predict the Cost of Living Index based upon Restaurant Price Index using a 95% level of confidence.

  1. Write the reqression equation.
  2. Discuss the statistical significance of the model as a whole using the appropriate regression statistic at a 95% level of confidence.
  3. Discuss the statistical significance of the coefficient for the independent variable using the appropriate regression statistic at a 95% level of confidence.
  4. Interpret the coefficient for the independent variable.
  5. What percentage of the observed variation in income is explained by the model?
  6. Predict the value of the cost of living index using this regression model with a Restaurant Price Index of 110.00

Create a table and compare the preceding three simple linear regression models to determine which model is the preferred model. Use the Significance F values, p-values for independent variable coefficients, R-squared or Adjusted R-squared values (as appropriate), and standard errors to explain your selection.

Data from Excel File

City

Cost of Living Index

Rent Index

Groceries Index

Restaurant Price Index

Local Purchasing Power Index

Luanda, Angola

150.82

138.03

171.77

121.3

50.27

Basel, Switzerland

145.92

62.61

139.31

134.41

144.12

Lausanne, Switzerland

140.88

57.61

144.89

117.3

165.05

Perth, Australia

138.97

72.08

126.03

131.02

112.02

Bern, Switzerland

136.3

57.31

133.11

124.59

158

Sydney, Australia

135.92

93.61

125.27

113.28

116.18

Solutions

Expert Solution

a)

Using Excel

data -> data analysis -> regression

SUMMARY OUTPUT
Regression Statistics
Multiple R 0.271149
R Square 0.073522
Adjusted R Square -0.1581
Standard Error 6.299925
Observations 6
ANOVA
df SS MS F Significance F
Regression 1 12.59825 12.59825 0.317424 0.603245
Residual 4 158.7562 39.68906
Total 5 171.3545
Coefficients Standard Error t Stat P-value Lower 95% Upper 95%
Intercept 117.0995 43.32936 2.702543 0.053952 -3.20214 237.4011
Restaurant Price Index 0.197079 0.349802 0.563404 0.603245 -0.77413 1.168284

y^ = 117.0995 + 0.1971 x

b)

F = 0.3174

p-value = 0.6032

since p-value > alpha

we fail to reject the null hypothesis

the model is not significant

c)

p-value of independent variable is also same 0.6032

since p-value > alpha

we fail to reject the null hypothesis

we conclude that the independent variable is not significant

d)

when restaurant price index increase by 1, on average cost of living index increase by 0.1971

e)

R^2 = 0.0735

hence 7.35 %

f)

x = 110

y^ = 117.0995 + 0.1971 *110

= 138.7805


Related Solutions

            Develop a simple linear regression model to predict the price of a house based upon...
            Develop a simple linear regression model to predict the price of a house based upon the living area (square feet) using a 95% level of confidence.             Write the reqression equation             Discuss the statistical significance of the model as a whole using the appropriate regression statistic at a 95% level of confidence.              Discuss the statistical significance of the coefficient for the independent variable using the appropriate regression statistic at a 95% level of confidence.             Interpret the...
. Develop a simple linear regression model to predict a person’s income (INCOME) based upon their...
. Develop a simple linear regression model to predict a person’s income (INCOME) based upon their years of education (EDUC) using a 95% level of confidence. a. Write the reqression equation. b. Discuss the statistical significance of the model as a whole using the appropriate regression statistic at a 95% level of confidence. c. Discuss the statistical significance of the coefficient for the independent variable using the appropriate regression statistic at a 95% level of confidence. d. Interpret the coefficient...
Develop a simple linear regression model to predict a person’s income (INCOME) based on their age...
Develop a simple linear regression model to predict a person’s income (INCOME) based on their age (AGE) using a 95% level of confidence. a. Write the regression equation. Discuss the statistical significance of the model as whole using the appropriate regression statistic at a 95% level of confidence. Discuss the statistical significance of the coefficient for the independent variable using the appropriate regression statistic at a 95% level of confidence. Interpret the coefficient for the independent variable. What percentage of...
1.Develop a multiple linear regression model to predict the price of a house using the square...
1.Develop a multiple linear regression model to predict the price of a house using the square feet of living area, number of bedrooms, and number of bathrooms as the predictor variables     Write the reqression equation.      Discuss the statistical significance of the model as a whole using the appropriate regression statistic at a 95% level of confidence. Discuss the statistical significance of the coefficient for each independent variable using the appropriate regression statistics at a 95% level of confidence....
Question 4 A simple linear regression model was used in order to predict y, duration of...
Question 4 A simple linear regression model was used in order to predict y, duration of relief from allergy, from x, dosage of medication. A total of n=10 subjects were given varying doses, and their recovery times noted. Here is the R output. summary(lmod4) ## ## Call: ## lm(formula = y ~ x) ## ## Residuals: ##     Min      1Q Median      3Q     Max ## -3.6180 -1.9901 -0.4798 2.2048 3.7385 ## ## Coefficients: ##             Estimate Std. Error t value Pr(>|t|)    ## (Intercept)...
You used multiple linear regression analysis to predict community reintegration (Reintegration to Normal Living Index RNLI;...
You used multiple linear regression analysis to predict community reintegration (Reintegration to Normal Living Index RNLI; interval scale) from depression (Geriatric Depression Scale or GDS; interval scale) and balance (Berg balance scale; interval scale) in a sample of 200 individuals with stroke. The results are as follows:                                         Model Summary R R Square Adjusted R Square Std. Error of the Estimate .670 .449 .431 14.40081 Predictors: (Constant), berg, depression                                                            ANOVA(c) Sum of Squares df Mean Square F Sig. Regression 10156.489 2...
A simple linear regression model based on 26 observations. The F-stat for the model is 6.45...
A simple linear regression model based on 26 observations. The F-stat for the model is 6.45 and the standard error for the coefficient of X is 0.2. MSR = 54.75 Complete an ANOVA table. Find the t-stat and the coefficient of X. Find R2.
Use Excel to develop a multiple regression model to predict Cost of Materials by Number of...
Use Excel to develop a multiple regression model to predict Cost of Materials by Number of Employees, New Capital Expenditures, Value Added by Manufacture, and End-of-Year Inventories. Locate the observed value that is in Industrial Group 12 and has 7 employees. Based on the model and the multiple regression output, what is the corresponding residual of this observation? Write your answer as a number, round to 2 decimal places. SIC Code No. Emp. No. Prod. Wkrs. Value Added by Mfg....
I have conducted a linear regression model to predict student scores on an exam based on...
I have conducted a linear regression model to predict student scores on an exam based on the number of hours they studied. I get a coefficient (slope) of +2.5 for the variable of hours studied. The pvalue for this coefficient is 0.45 and the 95% confidence interval is [-2.5, +7]. Which of the following conclusions CANNOT be drawn from these results? At an alpha of 0.05, we can say that the effect of hours studied on exam score is significant...
1. In the iris data, build a linear regression model to predict Sepal.Length based on both...
1. In the iris data, build a linear regression model to predict Sepal.Length based on both Petal.Length and Species. a. Calculate the regression equation, including the interaction. b. From this equation, you should be able to find 3 regression lines (one for each Species). Interpret each of the 3 slopes of the lines in the context of the problem. Remember that both numerical variables are measured in centimeters. c. Plot the 3 regression lines in a scatterplot of Sepal.Length vs....
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT