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• Comparing Two Regression Tests
• Independent T-test
• Nested ANOVA Test
• Linear Regression Test
• Three-way ANOVA Test
• Step-wise Linear Regression

## Comparing Two Regression Tests

Null hypothesis: there is no relationship between female mass and egg number in species 1
Alternative hypothesis: there is relationship between female mass and egg number in species1
Sum of Squares df Mean Square F P-value
Regression 41.0379 1 41.0379 72.2735 .000
Residual 9.6528 17 0.5678
Total 50.6907 18

The ANOVA table above shows that there is a strong evidence to reject the null hypothesis as the p value is less than 0.05. Therefore, we conclude that there is a linear relationship between the female mass and egg number for species 1.

 coefficient Std. Error t P-value (Constant) 4.1292 1.4409 2.8656 0.0107 Female Mass 0.2011 0.0237 8.5014 .0000

The regression equation is given as . The test shows that the variable (female mass) is significant at 5% level of significance. Null hypothesis: there is relationship between female mass and egg number in species2

Alternative hypothesis: there is no relationship between female mass and egg number in species2

Linear regression test

Sum of Squares df Mean Square F P-value
Regression 105.1745 1 105.1745 154.1147 .000
Residual 30.7099 45 0.6824
Total 135.8844 46

The ANOVA table above shows that there is a strong evidence to reject the null hypothesis as the p value is less than 0.05. Therefore, we conclude that there is a linear relationship between the female mass and egg number for species 2

 Coefficient Std. Error t P-value (Constant) 4.9654 0.9759 5.0880 0.000 Female Mass 0.1875 0.0151 12.4143 0.000

The regression equation is given as . The test shows that the variable (female mass) is significant at 5% level of significance ## Independent T-test

This is used to test the predicted values and species (grouping variable i.e. 1 and 2). The result is displayed below:

 Species N Mean Std. Deviation Std. Error Mean 1 19 61.0065954 6.94321843 1.59288355 2 47 63.8909508 7.11113481 1.03726562

The descriptive statistics show that Specie 1 has (N=19, mean = 61.01, SD = 6.94) while specie 2 has (N = 47, mean = 63.89, SD = 7.11).

Null hypothesis: the true difference between specie 1 and specie 2 is zero
Alternative hypothesis: the true difference between the groups is diff from zero
 F t df Sig. (2-tailed) Mean Difference .280 -1.502 64 .138 -2.88435533

The test shows that there is no sufficient evidence to reject the null hypothesis as the p-values is greater than 0.05, therefore, we conclude that the true difference between the group mean is zero i.e. the mathematical relationship between female mass and egg number is the same in both species.

## Nested ANOVA Test

Null hypothesis: Polycarp General Hospital (Smyrna) requires few number of days post op before release than Saint Genesius General Hospital (Rome)

Alternative hypothesis: Saint Genesius General Hospital (Rome) requires few number of days post op before release than Polycarp General Hospital (Smyrna)

The appropriate test for this question is the Nested ANOVA test, because we need to compare the means of both hospital to determine the better choice based on getting patient home sooner.

Nested ANOVA
Symra G.H Rome G.H
Mean 30.97143 28.37143
Variance 4.057143 4.584679
Observations 70 70
df 69 69
F 0.884935
P(F<=f) one-tail 0.306529
F Critical one-tail 0.671141

The table above shows the result for the nested ANOVA test. The mean and variance of Smyra General Hospital is 30.97, 4.057 and 28.37, 4.584 for Rome General Hospital respectively. The test statistic is 7.399, critical value is 1.977, p-value = 0.3065 This implies that we do not have strong evidence to reject the null hypothesis and conclude that Saint Polycarp General Hospital (Smyrna) requires few number of days post op before release than Genesius General Hospital (Rome). The best choice for patients who want to get home sooner is Saint Polycarp General Hospital (Smyrna).

Test for Normality for Smyra General Hospital
 Smyra G.H Kolmogorov-Smirnov Shapiro-Wilk Statistic df Sig. Statistic df Sig. .158 70 .000 .964 70 .042 The Kolmogorov-Smirnov test shows that the test is significant, therefore the data for Symra General Hospital is statistically significance, i.e. normality assumption test is valid.

##### Normality test for Rome General Hospital The Kolmogorov-Smirnov test shows that the test is significant, therefore, the data for Rome General Hospital is statistically significance, i.e. normality assumption test is valid.
The test also shows that the variance of the two test are equal.

## Linear Regression Test

Null hypothesis: there is no linear relationship between light and depth
Alternative hypothesis: there is a linear relationship between light and depth
The appropriate test for this question is the linear regression test. The reason for this test is because we are interested in the relationship between the two variable (light and depth).
Sum of Squares df Mean Square F P-value
Regression 1112490.230 1 1112490.23 14.766 .002
Residual 1130146.277 15 75343.085
Total 2242636.507 16

The ANOVA table above shows that there is a strong evidence to reject the null hypothesis as the p value is less than 0.05. Therefore, we conclude that there is a linear relationship between the light and depth and also a good fit between the two data (light and depth)

 coefficient Std. Error t P-value (Constant) 660.70 139.247 4.745 .000 Depth -52.218 13.589 -3.843 .002
The best regression equation is given as . The test shows that the variable (depth) is significant at 5% level of significance. ## Three-way ANOVA Test

Null hypothesis: there is no relationship between the independent variable (elevation feet, season and peaks) interactions

Alternative hypothesis: there is relationship between the independent variable (elevation feet, season and peaks) interactions

The most appropriate test for this question is the three way Anova test. This is because there are three different variables of interest and we want to know the interaction effect between the three variable. To determine the whether we have a statistically significant three-way interaction, we need to consult the “elevation”, “season”, “peaks” row in the Test of Between-subjects effect table as shown below.

Source Type III Sum of Squares df Mean Square F Sig. Partial Eta Squared
Corrected Model 117142.996a 35 3346.943 1367.016 .000 .997
Intercept 195261.735 1 195261.735 79752.179 .000 .998
Elevationfeet 31883.951 2 15941.976 6511.298 .000 .989
Season 3917.200 1 3917.200 1599.931 .000 .917
Peaks 34467.387 5 6893.477 2815.553 .000 .990
Elevationfeet * Season 644.600 2 322.300 131.639 .000 .646
Elevationfeet * Peaks 5645.555 10 564.556 230.586 .000 .941
Season * Peaks 34447.424 5 6889.485 2813.923 .000 .990
Elevationfeet * Season * Peaks 6136.878 10 613.688 250.653 .000 .946
Error 352.563 144 2.448
Total 312757.294 180
Corrected Total 117495.560 179
Dependent Variable: Weight
R Squared = .997 (Adjusted R Squared = .996)
The test above shows that there was a statistically significant (i.e. there is enough evidence to reject the null hypothesis) three way interaction between elevation, season and peaks . Since the test is statistically significant, we need to conduct the post hoc- test. The table is shown below
Multiple Comparisons
(I) Elevation(feet) (J) Elevation(feet) Mean Difference (I-J) Std. Error Sig. 95% Confidence Interval
Lower Bound Upper Bound
4000 6000 -16.1303* .28568 .000 -16.6950 -15.5657
8000 -32.6000* .28568 .000 -33.1647 -32.0353
6000 4000 16.1303* .28568 .000 15.5657 16.6950
8000 -16.4697* .28568 .000 -17.0343 -15.9050
8000 4000 32.6000* .28568 .000 32.0353 33.1647
6000 16.4697* .28568 .000 15.9050 17.0343
Dependent Variable: Weight
LSD

The post hoc test is used to confirm where the difference occurred between groups (elevation, season, peak). The table above shows that there is a statistically significant difference between all variable (elevation, season and peak) because the p-value of all the interaction are less at 0.05 at 5% level of significance. Null hypothesis: there is no linear relationship between AIC proteins and Age

Alternative hypothesis: there is linear relationship between AIC protein and Age
The most appropriate test for this question is the linear regression analysis, because we are interested in determining the relationship between two variables (AIC proteins and Age).
Sum of Squares df Mean Square F P-value
Regression 0.874 1 0.874 77.149 .000
Residual 0.589 52 0.011
Total 1.464 53
The ANOVA table above shows that there is a strong evidence to reject the null hypothesis as the p value is less than 0.05. Therefore, we conclude that there is a linear relationship between AIC protein and Age.
 coefficient Std. Error t P-value (Constant) 4.465 0.094 47.709 .000 Age 0.015 0.002 8.783 .000

The best regression equation is given as . The test shows that the variable (Age) is significant at 5% level of significance. The variable Age is significant because the p-value is less than 0.05, therefore we conclude that is it significant. The R square which is used to show the variability in AIC concentration is 0.597, which implies that the variable was able to explain about 60% of the AIC concentration.

Residual Plot Minimum Maximum Mean Std. Deviation N Predicted Value 5.0910 5.4635 5.2772 .12844 54 Residual -.20898 .20454 .00000 .10545 54 Std. Predicted Value -1.450 1.450 .000 1.000 54 Std. Residual -1.963 1.921 .000 .991 54 Dependent Variable: A1C

The table above shows the residual analysis of the dependent variable with mean and standard deviation of the residual and predicted value.

## Step-wise Linear Regression

Null hypothesis: there is no association between the dependent variable (Exact protein) and the independent variable (Protein 1, Protein 2, Protein 3, and Protein 4)

Alternative hypothesis: there is association between the dependent variable (Exact protein) and the independent variable (Protein 1, Protein 2, Protein 3, and Protein 4)

The appropriate test for this question is the stepwise linear regression (forward). This was chosen because we are interested in the final model among all model from the regression analysis.

Model R R Square Adjusted R Square Std. Error of the Estimate R Square Change Sig. F Change
1 .897a .805 .796 8.768 .805 .000
2 .966b .933 .927 5.251 .128 .000
3 .981c .962 .956 4.072 .029 .001

The table above shows the model summary, model 1 has an Adjusted R square of 0.796, model 2 has 0.927, while model 3 has an Adjusted R square of 0.962. It also shows that there is significant change in the Adjusted R square as the three model are statistically significant.

ANOVA
Model Sum of Squares df Mean Square F Sig.
1 Regression 7285.977 1 7285.977 94.782 .000b
Residual 1768.023 23 76.871
Total 9054.000 24
2 Regression 8447.343 2 4223.671 153.168 .000c
Residual 606.657 22 27.575
Total 9054.000 24
3 Regression 8705.803 3 2901.934 175.018 .000d
Residual 348.197 21 16.581
Total 9054.000 24
a. Dependent Variable: Exact[Protein]
b. Predictors: (Constant), Method 3[Protein]
c. Predictors: (Constant), Method 3[Protein], Method 1[Protein]
d. Predictors: (Constant), Method 3[Protein], Method 1[Protein], Method 4[Protein]

The ANOVA table above shows the model that are significant. This implies that model 1 (Method 3 protein) is statistically significant, model 2 (method 3 protein and method 1 protein) is statistically significant and model 3 (method 3 protein, method 1 protein and method 4 protein) is statistically significant.

 Model Unstandardized Coefficients Standardized Coefficients t Sig. B Std. Error Beta 1 (Constant) -106.133 20.447 -5.191 .000 Method 3[Protein] 1.968 .202 .897 9.736 .000 2 (Constant) -127.596 12.685 -10.059 .000 Method 3[Protein] 1.823 .123 .831 14.814 .000 Method 1[Protein] .348 .054 .364 6.490 .000 3 (Constant) -124.200 9.874 -12.578 .000 Method 3[Protein] 1.357 .152 .619 8.937 .000 Method 1[Protein] .296 .044 .310 6.784 .000 Method 4[Protein] .517 .131 .284 3.948 .001 a. Dependent Variable: Exact[Protein]

The regression equation of the final model is given as Residual plot for the predict value 