The dataset shows results for decathlon competition. This competition consists of 10 different disciplines.
- 100m (unit: seconds)
- Long jump (unit: metres)
- Shot put (unit: metres)
- High jump (unit: metres)
- 400 m (unit: seconds)
- 100 m hurdles (unit: seconds)
- Discus (unit: metres)
- Pole vault (unit: metres)
- Javelin (unit: metres)
- 1500 m (unit: seconds)
All columns with results for these disciplines consist of quantitative variables.
Also ‘Points’ are quantitative variables.
‘Rank’ and ‘Competition’ are both qualitative variables.
For ‘Competition’ we are having two values: ‘Decastar’ and ‘OlympicG’. The Decastar is an an- nual event and the Olympic is every 4 years, so we expect players in the OlympicG competition to have slightly better results (we are only guessing now).
The boxplot method shows us the groups of numerical data through their quarterlies. The bold line indicates the median, so the value above which we find 50% of data and below which we find another 50% of data. The upper bound of the box indicates the 3rd quartile and below the bound of the box the 1st quartile. Outliers are marked as dots.The median during the OlympicG competition is lower than for Decastar competition. The distance between 75% and 25% is lower for OlympicG competition.Whole range for Decastar competition is 11.65 - 10.75 = 0.9 seconds and for OlympicG: 11.4 - 10.5 = 0.9 seconds, however for OlympicG the range between second and third quarter is tighter. For OlypicG we can also notice one outlier with value ~ 10.4.
| Cell Contents |
|---|
| N |
| Chi-square contribution |
| N / Row Total |
| N / Col Total |
| N / Table Total |
Total Observations in Table: 41
| mydata$Competition
| mydata$X100m11 | Decastar | OlympicG | Row Total |
|---|---|---|---|
| faster | 2 | 19 | 21 |
| 3.259 | 1.513 | ||
| 0.095 | 0.905 | 0.512 | |
| 0.154 | 0.679 | ||
| 0.049 | 0.463 | ||
| --------------- | ----------- | ----------- | ----------- |
| slower | 11 | 9 | 20 |
| 3.422 | 1.589 | ||
| 0.550 | 0.450 | 0.488 | |
| 0.846 | 0.321 | ||
| 0.268 | 0.220 | ||
| --------------- | ----------- | ----------- | ----------- |
| Column Total | 13 | 28 | 41 |
| 0.317 | 0.683 | ||
| --------------- | ----------- | ----------- | ----------- |
Statistics for All Table Factors
Chiˆ2 = 9.783628 d.f. = 1 p = 0.
Chiˆ2 = 7.796182 d.f. = 1 p = 0.
The marginal probabilities are giving us an idea about the proportion between columns (num- ber of people in Decastar/OlympicG competition) and between rows (number of people with result better/worse than 11 seconds on 100 meters). We can see that there are more people in OlympicG competition (68.3%) than in Decastar (31.7%). The proportion between results better than 11 seconds and worse is more or less equal (51.2% and 48.8%). However, when we check cross table values we can see that in the Decastar competition there are only 2 people marked as ‘faster’ and 11 marked as ‘slower’. The Conditional probability P(‘faster’|Decastar) = 0.154 and P(‘slower’|Decastar) = 0.846. For the OlympicG competition there are more people marked as ‘faster’ (19 people) than those marked as ‘slower’ (9 people). The Conditional probability here is:
P(‘faster’|OlympicG) = 0.679 and P(‘slower’|OlympicG) = 0.321. Also when we read conditional probabilities P(Olympic|‘faster’) = 0.905 and P(Decastar|‘faster’) = 0.095 We can notice that there are faster people in OlympicG competition.
In my opinion this disproportion suggests that these two variables: Competition and categorical result on 100 meters are dependent.
The p-values for both Chi-squared tests (with and without Yates’ correction) are very small. We should reject our Ho hypothesis. That means the variables ”Competition” and ”X100m11” are dependent.
Shapiro-Wilk normality test
data: mydata$X100m W = 0.9818, p-value = 0.
Shapiro-Wilk normality test
data: mydata$Long.jump W = 0.98763, p-value = 0.
Shapiro-Wilk normality test
data: mydata$Shot.put W = 0.9884, p-value = 0.
Shapiro-Wilk normality test
data: mydata$High.jump W = 0.93734, p-value = 0.
Shapiro-Wilk normality test
data: mydata$X400m W = 0.95714, p-value = 0.
Shapiro-Wilk normality test
data: mydata$X110m.hurdle W = 0.93087, p-value = 0.
Shapiro-Wilk normality test
data: mydata$Pole.vault W = 0.97003, p-value = 0.
Shapiro-Wilk normality test
data: mydata$Javeline W = 0.97106, p-value = 0.
Shapiro-Wilk normality test
data: mydata$X1500m W = 0.93652, p-value = 0.
Shapiro-Wilk normality test
data: mydata$Points W = 0.95584, p-value = 0.
By looking on the graph and checking if p-value in shapiro test is higher than 0.05, we can say that:
Variables:
- X100m
- Long.jump
- Shot.put
- X400m
- Pole.vault
- Javeline
- Points
follow Normal distribution.
Variables:
-
High.jump
-
X110m.hurdle
-
X1500m
don’t follow Normal distribution.
H0 - There is no statistically significant difference between the samples
Welch Two Sample t-test
data: d1 and d t = 0.18568, df = 67.145, p-value = 0. alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -0.6817320 0. sample estimates: mean of x mean of y 5.064934 4.
We do not reject H0 hypothesis as p-value is greater than 0.
Two Sample t-test
data: d1 and d t = 18.83, df = 98, p-value < 2.2e- alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 3.666683 4. sample estimates: mean of x mean of y
We reject the H0 hypothesis as the p-value is smaller than 0.05. There is significant difference
Welch Two Sample t-test
data: d2 and d t = 10.634, df = 68.385, p-value = 3.884e- alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 3.272819 4. sample estimates: mean of x mean of y 4.9950070 0.
We reject the H0 hypothesis as the p-value is smaller than 0.05. There is a significant difference.
We can see that for tests where distributions have different means we always had to reject the H hypothesis and accept that there is significant difference between distributions.
Welch Two Sample t-test
data: mydata$X100m[mydata$Competition == "Decastar"] and␣ ,→mydata$X100m[mydata$Competition == "OlympicG"] t = 3.2037, df = 22.168, p-value = 0. alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 0.09164794 0. sample estimates: mean of x mean of y 11.17538 10.
We should reject the H0 hypothesis as the p-value is smaller than 0.05. There is a significant difference.
Welch Two Sample t-test
data: mydata$X400m[mydata$Competition == "Decastar"] and␣ ,→mydata$X400m[mydata$Competition == "OlympicG"] t = 0.05771, df = 32.106, p-value = 0. alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -0.6858299 0. sample estimates: mean of x mean of y 49.63 49.
We accept the H0 hypothesis as p-value is higher than 0.05. There is no significant difference.
Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1
Warning message in Boxplot.default(mf[[response]], x, id = list(method = id.method, : “NAs introduced by coercion”
As the p-value is less than the significance level 0.05, we can conclude that at least one distribution is different
2.2.1 ANOVA assumptions:
- For each population, the response variable is normally distributed (checked by Histogram, Q-Q plot or Shapiro-Wilk hypothesis test).
Shapiro-Wilk normality test
data: residuals(AnovaModel.1) W = 0.99309, p-value = 0.
The p-value is greater than 0.05, the assumption is valid.
- The variance of the response variable is the same for all of the populations (checked by Levene’s Test or Breusch Pagan Test).
studentized Breusch-Pagan test
data: AnovaModel. BP = 5.4666, df = 2, p-value = 0.
The p-value is greater than 0.05, the assumption is valid.
- The observations must be independent (checked by Durbin Watson test).
Durbin-Watson test
data: AnovaModel. DW = 1.9372, p-value = 0. alternative hypothesis: true autocorrelation is not 0
The p-value is greater than 0.05, the assumption is valid.
2.3.1 Questions:
- How does age influence on the risk factors associated with diabetes?
We were testing the hypothesis that there is no difference in means between groups. For the tests where this hypothesis was rejected (p-value smaller than 0.05) we can say that tested variable influences Age.
Variables related with age are:
- Pregnancies
- Glucose
- BloodPressure
- SkinThickness
- BMI
Insulin and DiabetesPedigreeFunction variables have no influence on age.
- Which of the risk factors are related with diabetes?
We were testing the hypothesis that there is no difference in means between groups. For the tests where this hypothesis was rejected we can say that the tested variable is related with diabetes.
Variables related with diabetes are:
- Pregnancies
- Glucose
- SkinThickness
- Insulin
- BMI
- DiabetesPedigreeFunction
Only the BloodPressure variable is not related.
- Detail the results of Two-Way ANOVA considering “Blood Pressure” as dependent variable, and the age groups and the indicator of diabetes as independent variables. Analyze the interaction term of two factors.
2.3.2 ANOVA assumptions
The observations within each sample must be independent (Durbin Watson).
Durbin-Watson test
data: model20 DW = 1.9684, p-value = 0.6611 alternative hypothesis: true autocorrelation is not 0
The populations from which the samples are selected must be normal (Shapiro test).
Shapiro-Wilk normality test
data: residuals(model20) W = 0.81264, p-value < 2.2e-16
The populations from which the samples are selected must have equal variances (Breusch Pa- gan test)
studentized Breusch-Pagan test
data: model20 BP = 9.7546, df = 5, p-value = 0.0825
The normality assumption is not valid but it’s the least important assumption.
There is no interaction.
Age:Outcome score is not significant.
We received only significant score for Age variable what means that Age is related with Blood- Pressure
- Analyze the distribution of Insulin variable. What would you recommend to fit an ANOVA model on Insulin levels?
I would recommend checking the dataset and removing errors and outliers. The distribution should be more similar to Normal distribution.
The highest correlation with 1500m is for 400m, so we will create simple linear regression with this feature.
Call: lm(formula = X1500m ~ X400m, data = train)
Residuals: Min 1Q Median 3Q Max -17.749 -6.411 -2.385 3.596 32.152
Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 34.105 87.982 0.388 0.70102 X400m 4.901 1.768 2.772 0.00949 **
Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 10.66 on 30 degrees of freedom Multiple R-squared: 0.2039, Adjusted R-squared: 0.1773 F-statistic: 7.683 on 1 and 30 DF, p-value: 0.009485
We can see that the decathlon$“400m” was marked by two stars. That means that variable is significant but not very strong (we didn’t receive three stars, which is a maximum).
The Residual standard error is equal 10.66, comparing it to the mean of the decathlon“1500m” (the variable we want to predict) which is equal to 277.9, we can say that the model is already relatively (quite) good. The Multiple R-squared is equal to 0.2039 which means that 20% of the variation of the response variable can be explained by using the decathlon$“400m” variable as the independent variable.
Call: lm(formula = X1500m ~ X400m + X100m + Long.jump + Shot.put + High.jump + X110m.hurdle + Discus + Pole.vault + Javeline + Competition, data = train)
Residuals: Min 1Q Median 3Q Max -19.3823 -3.4460 -0.2813 3.6758 21.2280
Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -174.3305 121.2106 -1.438 0.16510 X400m 7.7429 2.1107 3.668 0.00143 ** X100m 20.9493 11.2484 1.862 0.07660. Long.jump -0.9703 8.2885 -0.117 0.90792 Shot.put 1.2087 3.6820 0.328 0.74594 High.jump -9.8716 24.8265 -0.398 0.69492 X110m.hurdle -1.7192 5.1623 -0.333 0.74241 Discus 0.8561 0.8466 1.011 0.32341 Pole.vault 8.7085 7.1380 1.220 0.23598 Javeline 0.6821 0.4153 1.642 0.11543 Competition -6.8212 5.3146 -1.283 0.21330
Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 9.627 on 21 degrees of freedom Multiple R-squared: 0.5454, Adjusted R-squared: 0.3289 F-statistic: 2.519 on 10 and 21 DF, p-value: 0.03575
Only 400m and (a little bit) 100m results seem to be significant.
We can check if we meet assumptions
3.3.1 Normality
Shapiro-Wilk normality test
data: residuals(RegModel.1) W = 0.96362, p-value = 0.3439
We accept the H0 hypothesis (p-value higher than 0.05). The error term follows a Normal distri- bution. The normality assumption is valid.
3.3.2 Homogenity of Variance
studentized Breusch-Pagan test
data: RegModel.1 BP = 10.496, df = 10, p-value = 0.3981
We accept the H0 hypothesis as the p-value is higher than 0.05. The hypothesis is about homo- geneity of variance.
3.3.3 The independence of errors
Durbin-Watson test
data: RegModel.1 DW = 2.1561, p-value = 0.6798 alternative hypothesis: true autocorrelation is not 0
There are no autocorrelations in the dataset.
H0 is accepted as p-value is higher than 0.05.
The errors are independent
3.3.4 Multicollinearity
We can see a correlation between Shot.put and Discus.
The highly correlated variables can affect results. So we need to delete one of the correlated vari- ables.
Shot.put variable has the highest vif score, so that is the variable we should delete at the beginning.
Call: lm(formula = X1500m ~ X400m + X100m + Long.jump + High.jump + X110m.hurdle + Discus + Pole.vault + Javeline + Competition,
data = train)
Residuals: Min 1Q Median 3Q Max -19.4759 -3.3426 -0.4669 3.3364 21.5818
Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -173.23376 118.68218 -1.460 0.15852 X400m 7.69448 2.06238 3.731 0.00116 ** X100m 21.36551 10.94777 1.952 0.06383. Long.jump 0.07842 7.49140 0.010 0.99174 High.jump -7.22619 23.00127 -0.314 0.75636 X110m.hurdle -1.75596 5.05535 -0.347 0.73163 Discus 1.04860 0.59831 1.753 0.09360. Pole.vault 9.75740 6.25223 1.561 0.13288 Javeline 0.62156 0.36452 1.705 0.10225 Competition -6.47193 5.10030 -1.269 0.21773
Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 9.43 on 22 degrees of freedom Multiple R-squared: 0.543, Adjusted R-squared: 0.3561 F-statistic: 2.905 on 9 and 22 DF, p-value: 0.0199
3.5.1 Normality
Shapiro-Wilk normality test
data: residuals(RegModel.2) W = 0.95948, p-value = 0.2659
We accept H0 hypothesis (p-value higher than 0.05). The error term follows a Normal distribution. The normality assumption is valid.
3.5.2 Homogenity of Variance
studentized Breusch-Pagan test
data: RegModel.2 BP = 8.0252, df = 9, p-value = 0.5316
We accept the H0 hypothesis as the p-value is higher than 0.05. The hypothesis is about homo- geneity of variance.
3.5.3 The independence of errors
Durbin-Watson test
data: RegModel.2 DW = 2.1766, p-value = 0.5921 alternative hypothesis: true autocorrelation is not 0
There are no autocorrelations in the dataset.
H0 is accepted as p-value is higher than 0.05.
The errors are independent
3.5.4 Multicollinearity
Start: AIC=151.62
X1500m ~ X400m + X100m + Long.jump + High.jump + X110m.hurdle + Discus + Pole.vault + Javeline + Competition
Df Sum of Sq RSS AIC
- Long.jump 1 0.01 1956.4 149.62
- High.jump 1 8.78 1965.2 149.76
- X110m.hurdle 1 10.73 1967.2 149.79 1956.4 151.62
- Competition 1 143.19 2099.6 151.88
- Pole.vault 1 216.59 2173.0 152.98
- Javeline 1 258.56 2215.0 153.59
- Discus 1 273.15 2229.6 153.80
- X100m 1 338.70 2295.1 154.73
- X400m 1 1237.83 3194.3 165.31
Step: AIC=149.62 X1500m ~ X400m + X100m + High.jump + X110m.hurdle + Discus + Pole.vault + Javeline + Competition
Df Sum of Sq RSS AIC
- High.jump 1 8.83 1965.3 147.76
- X110m.hurdle 1 10.74 1967.2 147.80 1956.4 149.62
- Competition 1 147.00 2103.4 149.94
- Pole.vault 1 223.04 2179.5 151.07
- Javeline 1 264.74 2221.2 151.68
- Discus 1 275.14 2231.6 151.83
- X100m 1 375.64 2332.1 153.24
- X400m 1 1341.20 3297.6 164.33
Step: AIC=147.76 X1500m ~ X400m + X100m + X110m.hurdle + Discus + Pole.vault + Javeline + Competition
Df Sum of Sq RSS AIC
- X110m.hurdle 1 10.33 1975.6 145.93 1965.3 147.76
- Competition 1 139.87 2105.1 147.97
- Pole.vault 1 256.86 2222.1 149.69
- Discus 1 274.34 2239.6 149.95
- Javeline 1 297.63 2262.9 150.28
- X100m 1 376.28 2341.5 151.37
- X400m 1 1334.38 3299.6 162.35
Step: AIC=145.93 X1500m ~ X400m + X100m + Discus + Pole.vault + Javeline + Competition
Df Sum of Sq RSS AIC
1975.6 145.93
- Competition 1 142.38 2118.0 146.16
- Pole.vault 1 247.59 2223.2 147.71
- Discus 1 305.12 2280.7 148.53
- Javeline 1 310.33 2285.9 148.60
- X100m 1 502.94 2478.5 151.19
- X400m 1 1421.08 3396.7 161.27
Call: lm(formula = X1500m ~ X400m + X100m + Discus + Pole.vault + Javeline + Competition, data = train)
Coefficients: (Intercept) X400m X100m Discus Pole.vault Javeline -201.0539 7.4434 21.6470 1.0198 9.9059 0.6564 Competition -5.9655
The lowest AIC value the better. From the output of step() function we can see actual AIC score and AIC score when each variable will be deleted. Step by step we are deleting the variables to receive a model with as low AIC score as possible. When we see that deleting more variables will increase the AIC score we know that we already found the best multiple linear model.
Call: lm(formula = X1500m ~ X400m + X100m + Discus + Pole.vault + Javeline + Competition, data = train)
Residuals: Min 1Q Median 3Q Max -19.2804 -3.5023 -0.2449 2.8593 21.3204
Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -201.0539 95.4272 -2.107 0.045331 * X400m 7.4434 1.7553 4.241 0.000267 *** X100m 21.6470 8.5806 2.523 0.018377 * Discus 1.0198 0.5190 1.965 0.060621. Pole.vault 9.9059 5.5964 1.770 0.088914. Javeline 0.6564 0.3312 1.982 0.058612. Competition -5.9655 4.4443 -1.342 0.191573
Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 8.89 on 25 degrees of freedom
Multiple R-squared: 0.5386, Adjusted R-squared: 0.4278 F-statistic: 4.863 on 6 and 25 DF, p-value: 0.002038
3.7.1 Normality
Shapiro-Wilk normality test
data: residuals(RegModel.3) W = 0.95868, p-value = 0.2526
We accept the H0 hypothesis (p-value higher than 0.05). The error term follows a Normal distri- bution. The normality assumption is valid.
3.7.2 Homogenity of Variance
studentized Breusch-Pagan test
data: RegModel.3 BP = 7.3562, df = 6, p-value = 0.2892
We accept the H0 hypothesis as the p-value is higher than 0.05. The hypothesis is about homo- geneity of variance.
3.7.3 The independence of errors
Durbin-Watson test
data: RegModel.3 DW = 2.2015, p-value = 0.4967 alternative hypothesis: true autocorrelation is not 0
There are no autocorrelations in the dataset.
H0 is accepted as p-value is higher than 0.05.
The errors are independent
3.7.4 Multicollinearity
fit lwr upr
SEBRLE 270.9752 249.1711 292.7793
CLAY 281.8743 260.4247 303.3239
WARNERS 264.8520 244.5093 285.1946
BOURGUIGNON 276.3583 254.8415 297.8751
Karpov 263.2864 240.6217 285.9512
Warners 264.3165 244.4791 284.1538
Hernu 260.8604 240.0874 281.6333
Pogorelov 281.3147 261.6349 300.9945
Ojaniemi 262.6185 242.4955 282.7416
[512]: y <-test$'X1500m' y
291.7
301.5
278.1
291.7
278.11
278.05
264.35
287.63
275.71
Now we can compare fitted values with real values. We can also check that all fitted values are in
intervals (between lwr and upr value). That means our model works well.
RMSE value is smaller for the training set, which is correct as our model always should better fit
for data it was training on than for a new one.
RMSE for train set -> 7.85730086630928
RMSE for test set -> 14.3656946422001
In PCA we should not consider the X1500m variable, as we want to predict this variable. Also from the previous task we know that we also should not include Rank and Points variables in our model.
After eliminating X1500m, Points and Rank variables from PCA the variation in the first dimen- sion decreased and the variation in the second dimension increased.
eigenvalue percentage of variance cumulative percentage of variance
comp 1 3.3416485 33.416485 33.41649
comp 2 1.6910124 16.910124 50.32661
comp 3 1.2338861 12.338861 62.66547
comp 4 1.0807109 10.807109 73.47258
comp 5 0.7894027 7.894027 81.36661
comp 6 0.6134908 6.134908 87.50151
comp 7 0.4126831 4.126831 91.62834
comp 8 0.4052996 4.052996 95.68134
comp 9 0.2380949 2.380949 98.06229
comp 10 0.1937711 1.937711 100.00000
We can check the variation in each principal component, also the cumulative percentage of vari- ance, which means we know how many principal components we should leave to remain a specific percentage of variance.
We should find the elbow, so the number of principal components where eigenvalues are leveling off. On this graph it is 3 principal components, this allows us to keep almost 63% of variation.
Dim.1 Dim.2 Dim.3
X100m -0.80375792 0.2121185 0.36644639
Long.jump 0.71902293 -0.4315127 0.10507862
Shot.put 0.64115334 0.5592746 0.09576916
High.jump 0.55546011 0.3526622 0.39219320
X400m -0.64840502 0.5123973 -0.08437960
X110m.hurdle -0.73996769 0.2760499 -0.02235663
Discus 0.54841299 0.4854139 0.36783035
Pole.vault 0.03238867 -0.4217449 0.13236792
Javeline 0.29365964 0.4678044 -0.22591777
Competition 0.32092230 0.2270800 -0.84504163
We can check loading of each variable in each dimension. The absolute value of the variable is the value of loadings. The first dimension is an indicator mainly of ‘X100m’, ‘X110m.hurdle’ and ‘Long.jump’ variables. The second dimension is an indicator mainly of ‘Shot.put’ and ‘X400m’
variables. The third dimension is an indicator mainly of the ‘Competition’ variable.
4.0.1 Principal Component Regression
We can notice that between PCs there is almost no correlation.
Call: lm(formula = X1500m ~ PC1 + PC2 + PC3, data = decathlon)
Residuals: Min 1Q Median 3Q Max -23.291 -5.933 0.166 4.245 38.333
Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 279.0249 1.8411 151.555 <2e-16 *** PC1 -0.3440 1.0071 -0.342 0.735 PC2 0.8602 1.4158 0.608 0.547 PC3 2.1831 1.6574 1.317 0.196
Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 11.79 on 37 degrees of freedom Multiple R-squared: 0.05662, Adjusted R-squared: -0.01987 F-statistic: 0.7403 on 3 and 37 DF, p-value: 0.5348
There is no multicollinearity as we are using principals components (no correlation between PCs)
Assumptions
Normality
Shapiro-Wilk normality test
data: residuals(reg_pc_final) W = 0.94901, p-value = 0.06459
p-value > 0.05, test is valid
Homogenity of Variance
studentized Breusch-Pagan test
data: reg_pc_final BP = 2.6049, df = 3, p-value = 0.4566
p-value > 0.05, test is valid
The independence of errors
Durbin-Watson test
data: reg_pc_final DW = 1.8034, p-value = 0.3867 alternative hypothesis: true autocorrelation is not 0
p-value > 0.05, test is valid
All assumptions are valid.
RMSE = 11.1988259897779