a) draw a scatter diagram of number of sales calls and number of units sold
b) Estimate a simple linear regression model to explain the relationship between number of sales calls and number of units sold y=2.139x-1.760 Number of units sold=2.139Number of units sold-1.760 c) Calculate and interpret the coefficient of correlation r=0.853=0.9236 (There is strong correlation between two variables as its near 1) d) the coefficient of determination r2=0.853(The magnitude of the coefficient of determination indicates the proportion of variance in one variable, explained from knowledge of the second variable) e) the standard error of estimate S.E=0.3133(The standard error is the estimated standard deviation of a statistic) f) Conduct a test of hypothesis to determine whether the coefficient of correlation in the population is zero H0:β1=0 Ha:β1≠0
t=β1SE =6.826 p-value for df=9 and t=6.826:0.001
0.0001<0.05
Therefore null hypothesis is rejected
Hence coefficient of correlation is zero is rejected
Therefore there is significant relationship between number of sales calls and number of units sold. g) Construct and interpret confidence intervals and prediction intervals for the dependent variable, number of units sold.
Confidence interval:
(x-tsn,x+tsn)
Confidence interval for number of sales calls:
(x-tsn,x+tsn)
(0.924, 2.8612)
CALCULATIONS ON EXCEL
Regression Analysis | | | | | | | | | | | | | | | | r² | 0.853 | n | 10 | | | | | r | 0.924 | k | 1 | | | | | Std. Error | 8.412 | Dep. Var. | NUMBER OF UNITS SOLD | | | | | | | | | | ANOVA table | | | | | | | | Source | SS | df | MS | F |