# Solved – Multiple Linear Regression – p value equal to NA

I have the problem that my p-value is always NA.
I am making a Multiple Linear Regression in R. I have panel data for the period 2008-2018. One dependent variable (y) and 16 independent variables (x1-x16), z are the years (2008-2018).
Here are my inputs and results. I am using RStudio

``library(readxl) Risiko_Tool_Copy <- read_excel("C:/Users/debtaba1/Desktop/Risikotool/Risiko-Tool - Copy.xlsx") View(Risiko_Tool_Copy) fit <- lm(y ~ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 + x13 + x14 + x15 + x16 + z, data=Risiko_Tool_Copy) summary(fit) # show results  Call: lm(formula = y ~ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 +      x10 + x11 + x12 + x13 + x14 + x15 + x16 + z, data = Risiko_Tool_Copy)  Residuals: ALL 11 residuals are 0: no residual degrees of freedom!  Coefficients: (7 not defined because of singularities)               Estimate Std. Error t value Pr(>|t|) (Intercept) -8.941e+02         NA      NA       NA x1          -3.622e-01         NA      NA       NA x2           9.772e-03         NA      NA       NA x3          -4.864e+00         NA      NA       NA x4           1.809e+01         NA      NA       NA x5           2.086e+00         NA      NA       NA x6           1.123e+00         NA      NA       NA x7          -7.864e+00         NA      NA       NA x8           1.323e+01         NA      NA       NA x9          -9.386e-01         NA      NA       NA x10          1.165e-02         NA      NA       NA x11                 NA         NA      NA       NA x12                 NA         NA      NA       NA x13                 NA         NA      NA       NA x14                 NA         NA      NA       NA x15                 NA         NA      NA       NA x16                 NA         NA      NA       NA z                   NA         NA      NA       NA  Residual standard error: NaN on 0 degrees of freedom Multiple R-squared:      1, Adjusted R-squared:    NaN  F-statistic:   NaN on 10 and 0 DF,  p-value: NA ``

``structure(list(z = c(2008, 2009, 2010, 2011, 2012, 2013, 2014,  2015, 2016, 2017, 2018), y = c(0.956, 1.463, 0.457, 0.42, 0.57,  0.33, 0.2, 0.86, 1.14, 0.46, 0.67), x1 = c(2561.74, 2460.28,  2580.06, 2703.12, 2758.26, 2826.24, 2938.59, 3048.86, 3159.75,  3277.34, 3386), x2 = c(31719, 30569, 32137, 33673, 34296, 35045,  36287, 37324, 38370, 39650, 40883), x3 = c(7.8, 8.1, 7.7, 7.1,  6.8, 6.9, 6.7, 6.4, 6.1, 5.7, 5.2), x4 = c(3.88, 1.16, 1, 1.25,  0.88, 0.55, 0.16, 0.05, 0.01, 0, 0), x5 = c(82.3, 83, 83.9, 86.8,  89.8, 92.6, 95.5, 100, 106, 110.8, 116.3), x6 = c(92.5, 85.2,  97.1, 100.4, 96.7, 97.9, 99.3, 100, 100.2, 103.2, 103), x7 = c(2.6,  0.3, 1.1, 2.1, 2, 1.5, 0.9, 0.3, 0.5, 1.5, 1.8), x8 = c(97.8,  98.8, 100, 101.3, 102.5, 103.8, 105.4, 106.7, 108, 109.7, 110.9 ), x9 = c(80.76, 81.8, 81.75, 80.33, 80.52, 80.77, 81.2, 82.18,  82.52, 82.79, 82.88), x10 = c(615.8, 790.68, 1059.19, 1201.63,  1259.57, 867.14, 987.1, 971.9, 1095.28, 1084.83, 1119.42), x11 = c(95.98,  44.6, 79.36, 91.38, 98.92, 90.8, 98.71, 53.81, 36.98, 53.72,  60.42), x12 = c(95.98, 45.59, 77.93, 94.75, 107.24, 110.88, 110.91,  57.56, 37.61, 56.82, 66.87), x13 = c(4810.2, 5957.43, 6914.19,  5898.35, 7612.39, 9552.16, 9805.55, 10743, 11481.1, 12917.6,  10559), x14 = c(5601.9, 7507.04, 10128.12, 8897.81, 11914.37,  16574.45, 16934.85, 20774.62, 22188.94, 26200.77, 21588.09),      x15 = c(6052, 1070, 2848, 1107, 680, 699, 2277, 782, 478,      961, 1366), x16 = c(903.25, 1115.1, 1257.64, 1257.6, 1426.19,      1848.36, 2058.9, 2043.94, 2238.83, 2673.61, 2506.85)), row.names = c(NA,  -11L), class = c("tbl_df", "tbl", "data.frame")) ``
You will need at least $$m$$ observations to estimate $$m$$ parameters of a linear model. To do it reliably though will require much more, e.g. 50$$m$$ observations. In your case, $$m$$ is 18. So, technically you will need 18 observations. Practically, maybe 900, depending on the goal of your analysis.