 Solution

Use a common intercept and laboratory-specific slopes in the normal linear model. With $t_i$ denoting days and $I_{Bi},I_{Ci}$ the indicator variables for laboratories B and C, the design matrix has row $x_i^T=(1,t_i,t_i{I}_{Bi},t_i{I}_{Ci})$, andThe interaction terms change slopes, without introducing laboratory-specific intercepts. Thus the estimated daily growth rates arein millimetres per day, with fitted common intercept $0.6464$ mm. The residual degrees of freedom are $30-4=26$. The usual unbiased error variance estimate is the square of the residual standard error:The small difference from $5921.7/26\approx 227.758$ arises from rounding the standard error to two decimal places; the latter uses the more precise residual sum of squares supplied later. A maximum-likelihood estimate would instead divide that residual sum of squares by $30$.