Table 1.

Statistical tests and values

GraphType of testStatistical valuesStatistical values
a. Fig. 2A, preference for larger rewardz tests (50% preference; one-tailed); ANOVA (% difference in compared BSR frequencies)p < 0.001 (25–100% difference in BSR)*F(3,45) = 13.1, p < 0.001*p’s<0.001 (25-100% difference in BSR)*F(3,45)=13.1, p<0.001*
b. Fig. 2B, preference for larger rewardz tests (50% preference; one-tailed); ANOVA (individual frequency comparisons)p < 0.05 (all BSR comparisons)*F(10,118) = 6.995, p < 0.001*p’s<0.05 (all BSR comparisons)*F(10,118)=6.995, p<0.001*
c. Fig. 2C, preference for larger rewardz tests (50% preference; one-tailed); ANOVA (raw difference in compared BSR frequencies)all p < 0.001 (all BSR comparisons)*F(5,37) = 3.61, p = 0.011*all p’s<0.001 (all BSR comparisons)*F(5,37)=3.61, p=0.011*
d. Fig. 2D, number of trials completedANOVA (individual frequency comparisons)F(10,118) = 2.446, p = 0.011*F(10,118)=2.446, p=0.011*
e. Fig. 2E, preference for larger rewardz tests (50% preference; one-tailed); ANOVA (proportionate time × % difference in BSR frequency); ANOVA (proportionate time); ANOVA (% difference in BSR frequency)p ≤ 0.01 (all BSR comparisons)*F(27,279) = 1.346, p = 0.133F(9,279) = 5.838, p < 0.001*F(3,279) = 20.582, p < 0.001*p’s≤0.01 (all BSR comparisons)*F(27,279)=1.346, p=0.133F(9,279)=5.838, p<0.001*F(3,279)=20.582, p<0.001*
f. Fig. 2F, number of trials completedANOVA (proportionate time × % difference in BSR frequency); ANOVA (proportionate time); ANOVA (% difference in BSR frequency)F(27,279) = 1.242, p = 0.205F(9,279) = 6.320, p < 0.001*F(3,279) = 1.457, p = 0.259F(27,279)=1.242, p=0.205F(9,279)=6.320, p<0.001*F(3,279)=1.457, p=0.259
g. Fig. 3A, preference for larger rewardz tests (50% preference; one-tailed); ANOVA (difference in pellet number)p < 0.001 (all pellet comparisons)*F(3,39) = 8.63, p < 0.001*p’s<0.001 (all pellet comparisons)*F(3,39)=8.63, p<0.001*
h. Fig. 3B, preference for larger rewardz tests (50% preference; one-tailed); ANOVA (individual pellet comparisons within each comparison range)p < 0.001 (1v2, 2v3, 1v3, 2v4, 1v4, 2v5, 1v5)*F(10,118) = 7.00, p < 0.001*p’s<0.001 (1v2, 2v3, 1v3, 2v4, 1v4, 2v5, 1v5)*F(10,118)=7.00, p<0.001*
i. Fig. 3C, number of trials completedANOVA (individual pellet comparisons within each comparison range)F(10,109) = 5.40, p < 0.001*F(10,109)=5.40, p<0.001*
j. Fig. 3D, number of trials completedPearson correlationR 2 = 0.998, p < 0.0001*R2=0.998, p<0.0001*
k. Fig. 3E, preference for larger rewardz tests (50% preference; one-tailed); ANOVA (proportionate time × % difference in pellet number); ANOVA (proportionate time); ANOVA (% difference in pellet number)p ≤ 0.05 (0.0–0.1 proportionate time)*F(27,399) = 1.509, p = 0.056F(9,399) = 25.29, p < 0.001*F(3,399) = 1.91, p = 0.152p≤0.05 (0.0-0.1 proportionate time)*F(27,399)= 1.509, p=0.056F(9,399)= 25.29, p<0.001*F(3,399)= 1.91, p=0.152
l. Fig. 3F, number of trials completedANOVA (proportionate time × difference in pellet number); ANOVA (proportionate time); ANOVA (% difference in pellet number)F(27,399) = 0.878, p = 0.64F(9,399) = 112.136, p < 0.001*F(3,399) = 1.146, p = 0.349F(27,399)=0.878, p=0.64F(9,399)=112.136, p<0.001*F(3,399)=1.146, p=0.349
m. Fig. 4A, preference for 1 sugar pelletz test (50% preference; one-tailed); ANOVA (proportionate BSR frequency)p < 0.05 (0%, 50% BSR)*F(4,24) = 0.413, p = 0.80p’s<0.05 (0%, 50% BSR)*F(4,24)=0.413, p=0.80
n. Fig. 4B, preference for 2 sugar pelletsz test (50% preference; one-tailed); ANOVA (proportionate BSR frequency)p < 0.001 (0%, 25% BSR)*F(4,24) = 0.963, p = 0.449p’s<0.001 (0%, 25% BSR)*F(4,24)=0.963, p=0.449
o. Total pellets earnedANOVA (sugar pellet reward size × BSR reward size); ANOVA (sugar pellet reward size); ANOVA (BSR reward size)F(4,49) = 1.486, p = 0.253F(1,49) = 66.31, p < 0.001*F(4,49) = 1.037, p = 0.418F(4,49)=1.486, p=0.253F(1,49)=66.31, p<0.001*F(4,49)=1.037, p=0.418
p. Fig. 4C, preference for 1 sugar pelletz test (50% preference; one-tailed); ANOVA (proportionate time × proportionate BSR frequency); ANOVA (proportionate time); ANOVA (proportionate BSR frequency)all p < 0.05 (0.3–1.0 proportionate time)*F(36,249) = 0.928, p = 0.59F(9,249) = 3.987, p = 0.001F(4,249) = 0.82, p = 0.531all p’s<0.05 (0.3-1.0 proportionate time)*F(36,249)=0.928, p=0.59F(9,249)=3.987, p=0.001F(4,249)=0.82, p=0.531
q. Fig. 4D, preference for 2 sugar pelletsz test (50% preference; one-tailed); ANOVA (proportionate time × proportionate BSR frequency); ANOVA (proportionate time); ANOVA (proportionate BSR frequency)all p < 0.05 (0.6–1.0 proportionate time)*F(36,249) = 1.886, p = 0.005*F(9,249) = 10.60, p < 0.001*F(4,249) = 1.295, p = 0.314all p’s<0.05 (0.6-1.0 proportionate time)*F(36,249)=1.886, p=0.005*F(9,249)=10.60, p<0.001*F(4,249)=1.295, p=0.314
r. Fig. 4E, number of trials completed (1-pellet sessions)ANOVA (proportionate time × proportionate BSR frequency); ANOVA (proportionate time); ANOVA (proportionate BSR frequency)F(36,249) = 0.916, p = 0.608F(9,249) = 3.353, p = 0.004*F(4,249) = 1.538, p = 0.239F(36,249)=0.916, p=0.608F(9,249)=3.353, p=0.004*F(4,249)=1.538, p=0.239
s. Fig. 4F, number of trials completed (2-pellet sessions)ANOVA (proportionate time × proportionate BSR frequency); ANOVA (proportionate time); ANOVA (proportionate BSR frequency)F(36,249) = 1.360, p = 0.105F(9,249) = 4.673, p < 0.001*F(4,249) = 0.606, p = 0.664F(36,249)=1.360, p=0.105F(9,249)=4.673, p<0.001*F(4,249)=0.606, p=0.664