Monthly Return Comparison Between Cummins and Vanguard S&P 500 Index ETF

The following chart shows the Vanguard S&P 500 Index ETF (VOO) monthly returns on the x-axis and Cummins (CMI) on the y-axis. The regression line on the graph shows a steep slope, and the Pearson correlation value is 0.7. This value shows a very strong correlation between the monthly returns of the Vanguard S&P 500 Index ETF and Cummins. The p-value is significant at a 95% confidence interval. Cummins has returned 13.9% in the past year, while the Vanguard S&P 500 Index ETF has returned a -12.4%.  

Exhibit: Monthly Returns of Vanguard S&P 500 Index ETF and Cummins [June 2019 - October 2022]

   

The following command was used to create this graph:

> # Create a new Graph of $VOO and $CMI Monthly Returns as Percentages

> ggscatter(df1, x = 'VOO_Monthly_Return', y = 'CMI_Monthly_Return', 

+           add = "reg.line", conf.int = TRUE, 

+           cor.coef = TRUE, cor.method = "pearson",

+           xlab = "VOO ETF Monthly Returns (%)", ylab = "CMI Monthly +           Returns (%)")

A linear regression of the monthly returns of the Vanguard S&P 500 Index ETF and Cummins shows the beta (coefficient of Vanguard S&P 500 Index ETF) for Cummins' monthly returns compared to the Vanguard ETF. 

# Conduct the Linear Regression of the Monthly Returns Between $VOO and $CMI

lmVOOCMI = lm(CMI_Monthly_Return~VOO_Monthly_Return, data = VOOandCMI)

# Present the summary of the results from the linear regression

summary(lmVOOCMI)

Call:

lm(formula = CMI_Monthly_Return ~ VOO_Monthly_Return, data = VOOandCMI)

Residuals:

      Min        1Q    Median        3Q       Max 

-0.123268 -0.051037  0.004537  0.047062  0.115727 

Coefficients:

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

(Intercept)        0.005104   0.009226   0.553    0.583    

VOO_Monthly_Return 0.993327   0.160541   6.187 2.84e-07 ***

---

Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.05819 on 39 degrees of freedom

Multiple R-squared:  0.4954, Adjusted R-squared:  0.4824 

F-statistic: 38.28 on 1 and 39 DF,  p-value: 2.845e-07

Cummins has a beta of 0.99 (nearly 1). Given this beta, Cummins, on average, moves in line with the S&P 500 Index. The adjusted R-squared value of 0.48 shows that about 48% of Cummins' monthly returns can be attributed to the returns of the S&P 500 Index. The p-value of 2.84e-07 shows that the regression analysis results are significant at the 95% confidence interval.

The average monthly return for Cummins between June 2019 and October 2022 was 1.49%. A one-sample t-test shows that the monthly return falls between -1.05% and 4.04%. But, the p-value of 0.99 is much higher than 0.05. This t-test may not be significant in the 95% confidence interval.        

#

# One Sample t-test of $CMI average monthly returns.

# 

# Step 1: 

# Copy Dataframe Column CMI_Monthly_Return into a List 

#

CMI_Monthly_Return_Col <- c(VOOandCMI['CMI_Monthly_Return'])

# CMI_Monthly_Return_Col is a list object, but needs to be numeric

# for qqnorm to work.  

#

typeof(CMI_Monthly_Return_Col)

# Convert List Object into a Column of doubles as as.numeric and unlist

#

y_CMI_Monthly_Return_Col <- as.numeric(unlist(CMI_Monthly_Return_Col))

typeof(y_CMI_Monthly_Return_Col)

# Let’s check if the data comes from a normal distribution 

# using a normal quantile-quantile plot.

# Source: https://cran.r-project.org/web/packages/distributions3/vignettes/one-sample-t-test.html

Exhibit: Check if Cummins' Monthly Returns are Normally Distributed before doing a t-test

#

qqnorm(y_CMI_Monthly_Return_Col)

qqline(y_CMI_Monthly_Return_Col)

# Conduct a t-test to see if the population mean is 1.49% [0.0149]

#

t.test(y_CMI_Monthly_Return_Col, mu = .0149)

One Sample t-test

data:  y_CMI_Monthly_Return_Col

t = 0.0045069, df = 40, p-value = 0.9964

alternative hypothesis: true mean is not equal to 0.0149

95 percent confidence interval:

 -0.01057188  0.04048574

sample estimates:

 mean of x 

0.01495693