Question

e monthly sales (in hundreds of dollars) of a company are listed below Month Sales (S100s 12,254 13,557 14,436 13,378 16,490 19,690 17,887 18,557 19.665 18,578 20,578 23,575 an March May June Jul August September October November December a. Construct a time series plot. What type of pattern exists in the data? b. Use simple linear regression analysis to find for this time serie:s the parameters for the line that minimi zes MSE c. What is the sales forecast (in hundreds of dollars) for next month? Hint: Replace the months with numbers

Answer 1

_{Here we have indexed months from 1 to 12}

month Files Plots Packages Help Viewer [1] 1 2 3 4 5 6 7 8 9 10 11 12 Sales [1] 12254 13557 14436 13378 16490 19690 17887 18557 19665 18578 20578 23575 > plot (month,sales) > myfit-1m(sales-month) summary(myfit) call: 1m(formula Sales month) Residuals: -1899. 69 -402.99 64. 37 214.53 2744.43 Coefficients: Min 1Q Median Max Estimate Std. Error t value Pr ltlI (Intercept) 11647.4 month 804.9 14.471 4.93e-08 88R 109.4 8.074 1.09e-05** 883.0 signif. codes : o ' , 0.001 .**, o.01 .*' 0. 05 ..' 0.1 ' ' 1 Residual standard error: 1308 on 10 degrees of freedom Multiple R-squared 0.867, F-statistic: 65.2 on 1 and 10 DF, p-value: 1.086e-05 10 12 4 Adjusted R-squared: 0.8537

A) Time series plot of the given data is plotted in R and given above on the right hand side.

from the plot is clearly visible that there is linear pattern between time and sales i.e, as time sales increases

B) we have fitted linear regression model on the given data and parameters that minimize MSE (mean squared error) is given by:
regression line is :

sales = 11647.4 + 883*month

parameters

intercept = 11747.4

slope = 883

C) for next month i.e, month = 13 we can predict the sales in 100 $ by putting month = 13 in the given above regression line equation

sales= 11647.4 + 883*13 = 23126.79