Linear Resgression
Aritra Halder
2/17/2021
- The sales of a particular product along with the amount of money spent for its advertisement in ten different markets are recorded as follows. Advertisement Expenditure (in thousand Rs.) 16 17 20 19 26 15 18 12 15 22 Sales Volume(in ten thousand unit) 13.9 13.4 14.9 12.7 15.3 11.9 14.0 12.8 14.1 14.0 From this record calculate the following
- Increment in sales volume if advertisement expenditure is increased by ten thousand rupees
- Volume of sales if there is no investment in the advertisement
x <- c(16, 17, 20, 19, 26, 15, 18, 12, 15, 22)
y <- c(13.9, 13.4, 14.9, 12.7, 15.3, 11.9, 14.0, 12.8, 14.1, 14.0)
lmObj = lm(y~x)
summary(lmObj)##
## Call:
## lm(formula = y ~ x)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.2812 -0.3255 0.1771 0.4844 0.9187
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 10.58750 1.23763 8.555 2.69e-05 ***
## x 0.17292 0.06728 2.570 0.0331 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.8073 on 8 degrees of freedom
## Multiple R-squared: 0.4523, Adjusted R-squared: 0.3838
## F-statistic: 6.606 on 1 and 8 DF, p-value: 0.03312cat("Increase in sales volume when advertisement is increased by Rs 10000 is ",0.17292*10," in ten thousand units.")## Increase in sales volume when advertisement is increased by Rs 10000 is 1.7292 in ten thousand units.cat("\n The intercept ",10.58750," will be the sales without any advertisement.")##
## The intercept 10.5875 will be the sales without any advertisement.plot(x,y)
abline(lmObj,col="red")
14. The following data give the annual incomes (in thousands of dollars) and amounts (in thousands of dollars) of life insurance policies for eight persons. Annual income (x) 42 58 27 36 70 24 53 37 Life insurance (y) 150 175 25 75 250 50 250 100 (a) Calculate the least-squares regression line for these data. (b) Plot the points and the least-squares regression line on the same graph. (c) Calculate the 95% confidence intervals for B0 and B1, respectively. (d) Test the hypothesis H0 : B0 = 0 vs. Ha : B0 != 0 using the 0.05 level of significance. (e) Test the hypothesis H0 : B1 = 0 vs. Ha : B1 != 0 using the 0.05 level of significance. (f) Obtain a 90% prediction interval at x = 59 and interpret the result.
x <- c(42, 58, 27, 36, 70, 24, 53, 37)
y <- c(150, 175, 25, 75, 250, 50, 250, 100)
lmObj = lm(y~x)
summary(lmObj)##
## Call:
## lm(formula = y ~ x)
##
## Residuals:
## Min 1Q Median 3Q Max
## -33.06 -23.38 -10.39 15.57 67.13
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -84.1674 39.5586 -2.128 0.07746 .
## x 5.0384 0.8631 5.838 0.00111 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 36.16 on 6 degrees of freedom
## Multiple R-squared: 0.8503, Adjusted R-squared: 0.8254
## F-statistic: 34.08 on 1 and 6 DF, p-value: 0.001113confint(lmObj)## 2.5 % 97.5 %
## (Intercept) -180.963802 12.628926
## x 2.926612 7.150272plot(x,y)
abline(lmObj,col="red")
print("We fail to reject the null hypothesis that the intercept is equal to zero")## [1] "We fail to reject the null hypothesis that the intercept is equal to zero"print("We reject the null hypothesis that the slope is equal to zero ")## [1] "We reject the null hypothesis that the slope is equal to zero "predict(lmObj,data.frame(x=59),interval = "prediction",level = 0.9)## fit lwr upr
## 1 213.1007 134.0908 292.1105- The following data represent systolic blood pressure readings on 10 randomly selected females between ages 40 and 82. Age (x) 63 70 74 82 60 44 80 71 71 41 Systolic (y) 151 149 164 157 144 130 157 160 121 125
- Calculate the least-squares regression line for these data.
- Plot the points and the least-squares regression line on the same graph.
- Calculate the 95% confidence intervals for B0 and B1, respectively.
- Test the hypothesis H0 : B0 = 0 vs. Ha : B0 != 0 using the 0.05 level of significance.
- Test the hypothesis H0 : B1 = 0 vs. Ha : B1 != 0 using the 0.05 level of significance.
- Obtain a 95% prediction interval at x = 85 and interpret the result.
x <- c(63, 70, 74, 82, 60, 44, 80, 71, 71, 41)
y <- c(151, 149, 164, 157, 144, 130, 157, 160, 121, 125)
lmObj = lm(y~x)
summary(lmObj)##
## Call:
## lm(formula = y ~ x)
##
## Residuals:
## Min 1Q Median 3Q Max
## -28.8606 -0.8763 0.4071 5.9691 11.8835
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 96.4714 19.1861 5.028 0.00102 **
## x 0.7520 0.2867 2.622 0.03054 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 11.94 on 8 degrees of freedom
## Multiple R-squared: 0.4622, Adjusted R-squared: 0.395
## F-statistic: 6.877 on 1 and 8 DF, p-value: 0.03054plot(x,y)
abline(lmObj,col="red")
confint(lmObj)## 2.5 % 97.5 %
## (Intercept) 52.22817238 140.714632
## x 0.09071445 1.413206print("We reject the null hypothesis that the intercept is equal to zero")## [1] "We reject the null hypothesis that the intercept is equal to zero"print("We reject the null hypothesis that the slope is equal to zero ")## [1] "We reject the null hypothesis that the slope is equal to zero "predict(lmObj,data.frame(x=85),interval = "prediction")## fit lwr upr
## 1 160.388 128.7849 191.9912- The data given below are from a random sample of height (in inches) and weight (in pounds) of seven basketball players. Height (x) 73 83 77 80 85 71 80 Weight (y) 186 234 208 237 265 190 220
- Calculate the least-squares regression line for these data.
- Plot the points and the least-squares regression line on the same graph.
- Calculate the 95% confidence intervals for 0 and 1, respectively.
- Test the hypothesis H0 : B0 = 0 vs. Ha : B0 != 0 using the 0.05 level of significance.
- Test the hypothesis H0 : B1 = 0 vs. Ha : B1 != 0 using the 0.05 level of significance.
- Construct a 99% prediction interval for height equal to 90. Interpret the result and state any assumptions.
x <- c(73, 83, 77, 80, 85, 71, 80)
y <- c(186, 234, 208, 237, 265, 190, 220)
lmObj = lm(y~x)
summary(lmObj)##
## Call:
## lm(formula = y ~ x)
##
## Residuals:
## 1 2 3 4 5 6 7
## -5.727 -9.809 -4.560 8.816 10.774 8.690 -8.184
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -188.4761 62.0831 -3.036 0.02889 *
## x 5.2083 0.7902 6.591 0.00121 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 9.86 on 5 degrees of freedom
## Multiple R-squared: 0.8968, Adjusted R-squared: 0.8762
## F-statistic: 43.45 on 1 and 5 DF, p-value: 0.001207plot(x,y)
abline(lmObj,col="red")
confint(lmObj)## 2.5 % 97.5 %
## (Intercept) -348.065841 -28.886453
## x 3.177085 7.239429print("We reject the null hypothesis that the intercept is equal to zero")## [1] "We reject the null hypothesis that the intercept is equal to zero"print("We reject the null hypothesis that the slope is equal to zero ")## [1] "We reject the null hypothesis that the slope is equal to zero "predict(lmObj,data.frame(x=90),interval = "prediction",level = 0.99)## fit lwr upr
## 1 280.267 224.0032 336.5308- The following data relate to the prices (Y ) of five randomly chosen houses in a certain neighborhood, the corresponding ages of the houses (X1), and square footage (X2).
Price Y in thousands of dollars : 100,80,104,94,130 Age X1 in years : 1,5,5,10,20 Square Footage X2 in thousands of square feet : 1,1,2,2,3
- Fit a multiple linear regression model to the foregoing data.
- Estimate the error variance.
- Obtain an ANOVA table and test the hypothesis H0 : B1 = B2 = 0 vs. Ha : At least one of the Bi != 0, i = 1, 2. Use alpha = 0.05.
x1 <- c(1,5,5,10,20)
x2 <- c(1,1,2,2,3)
y <- c(100,80,104,94,130)
lmObj = lm(y~x1+x2)
summary(lmObj)##
## Call:
## lm(formula = y ~ x1 + x2)
##
## Residuals:
## 1 2 3 4 5
## 12.818 -5.665 -3.101 -11.204 7.153
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 66.1252 21.0132 3.147 0.0879 .
## x1 -0.3794 2.2212 -0.171 0.8801
## x2 21.4365 19.4544 1.102 0.3854
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 13.83 on 2 degrees of freedom
## Multiple R-squared: 0.7142, Adjusted R-squared: 0.4285
## F-statistic: 2.499 on 2 and 2 DF, p-value: 0.2858anova(lmObj)## Analysis of Variance Table
##
## Response: y
## Df Sum Sq Mean Sq F value Pr(>F)
## x1 1 724.17 724.17 3.7845 0.1911
## x2 1 232.33 232.33 1.2141 0.3854
## Residuals 2 382.70 191.35print("We fail to reject the null hypothesis.")## [1] "We fail to reject the null hypothesis."- A study is conducted to estimate the demand for housing (Y ) based on current interest rate (X1) and the rate of unemployment (X2). The data in the following table are obtained.
Units Sold : 65,59,80,90,100,105 Interest rate : 9.0,9.3,8.9,9.1,9.0,8.7 Unemployment rate : 10.0,8.0,8.2,7.7,7.1,7.2
- Fit a multiple linear regression model.
- Estimate the error variance.
- Test whether the model is significant. Use alpha = 0.05.
x1 <- c(9.0,9.3,8.9,9.1,9.0,8.7)
x2 <- c(10.0,8.0,8.2,7.7,7.1,7.2)
y <- c(65,59,80,90,100,105)
lmObj = lm(y~x1+x2)
summary(lmObj)##
## Call:
## lm(formula = y ~ x1 + x2)
##
## Residuals:
## 1 2 3 4 5 6
## 3.522 -8.477 -6.681 8.510 6.540 -3.414
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 653.469 185.861 3.516 0.0390 *
## x1 -53.523 20.988 -2.550 0.0839 .
## x2 -11.028 3.976 -2.774 0.0694 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 9.233 on 3 degrees of freedom
## Multiple R-squared: 0.8522, Adjusted R-squared: 0.7537
## F-statistic: 8.652 on 2 and 3 DF, p-value: 0.0568anova(lmObj)## Analysis of Variance Table
##
## Response: y
## Df Sum Sq Mean Sq F value Pr(>F)
## x1 1 819.20 819.20 9.6094 0.05330 .
## x2 1 655.88 655.88 7.6937 0.06935 .
## Residuals 3 255.75 85.25
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1print("We fail to reject the null hypothesis.")## [1] "We fail to reject the null hypothesis."