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Home/ Questions/Q 6962557
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T15:40:34+00:00 2026-05-27T15:40:34+00:00

I tried using manova() and can not seem to get the correct programming in.

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I tried using manova() and can not seem to get the correct programming in. I tried it this way (in a different post):

manova in R error message: length of 'dimnames' [1] not equal to array extent

the text question has to do with pedal rotation and initial speed being predictors of acceleration.

here is that data:

acc <- data.frame(Degrees = c("d5","d8","d10"), MPH10=c(0.35, 0.37, 0.32), 
MPH25=c(0.19, 0.28, 0.30), MPH40=c(0.14, 0.19, 0.29), MPH55=c(0.10, 0.19, 0.23))

acc

acc
  Degrees MPH10 MPH25 MPH40 MPH55
1     5  0.35  0.19  0.14  0.10
2     8  0.37  0.28  0.19  0.19
3     10  0.32  0.30  0.29  0.23

no idea what to do next.

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  1. Editorial Team
    Editorial Team
    2026-05-27T15:40:34+00:00Added an answer on May 27, 2026 at 3:40 pm

    The conventional answer would be:

    macc <- melt(acc, id.var="Degrees")
    lm(value ~ Degrees + variable, macc)
    anova(lm(value ~ Degrees + variable, macc))
    

    And all that remains is constructing a proper description of the results. (Notice that I used “+” instead of “*”). You get a nearly perfect answer when constructing a saturated model (one with no residuals) when using the interaction model:

    anova(lm(value ~ Degrees * variable, macc))
    

    You could have coded either or both of Degrees or MPH variables as numeric and gotten an unsaturated model.But it would still have added to the complexity of describing the result.

     acc <- data.frame(Degrees = c(5,8,10), MPH10=c(0.35, 0.37, 0.32), 
         MPH25=c(0.19, 0.28, 0.30), MPH40=c(0.14, 0.19, 0.29), MPH55=c(0.10, 0.19, 0.23))
     macc <- melt(acc, id.var="Degrees")
     anova(lm(value ~ Degrees * variable, macc))
    

    Using sub to remove the “MPH” from the character variables. I thought it would be necessary to use as.numeric(as.character()) on what I thought would be a factor variable, but the sub operation apparently stripped the factor attribute and just using as.numeric was sufficient.

    macc$speed <- as.numeric(sub("MPH", "", macc$variable))
    anova(lm(value ~ Degrees + speed, macc))
    # output omitted
    anova(lm(value ~ Degrees * speed, macc))
    #-------------------
        Analysis of Variance Table
    
    Response: value
                  Df   Sum Sq  Mean Sq F value   Pr(>F)    
    Degrees        1 0.016827 0.016827  16.904 0.003384 ** 
    speed          1 0.048735 0.048735  48.959 0.000113 ***
    Degrees:speed  1 0.006367 0.006367   6.396 0.035309 *  
    Residuals      8 0.007963 0.000995                     
    ---
    Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 
    
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