Does increased air temperature cause increased maximum height of a jump by a person straight up into the air? Assuming constant pressure, when the air is hotter, on a hot day, it will have less density and therefore cause a smaller fluid upthrust but also exert less air resistance.
So, on a hot day, other things being equal (humidity and performance by the body being ignored - Thanks GS), someone who jumps straight up will be able to jump less high due to the reduced fluid upthrust, but higher due to the reduced air resistance.
So my question is which effect will predominate, and therefore will a person be able to jump higher on a hot day or on a cold one, other things equal?
 A: Person Jumping is bad metric because people are messy machines and the tolerance you're working in will not detect any difference. In reality the person will probably jump a measurable amount higher on the hot day... because muscles and the respiratory system work better when the weather is warm. The weight difference from buoyancy could also be made up for by cutting off a few inches of a shoelace.
Hot Day is also a bad metric, because the air's likely to be more dense on the hot day than the cold day, not less: hot air can retain much more water and thus tends to be more dense.
Anyway, both effects are so small as to be negligible for a sample as messy as a human and an environmental variable as messy as weather.
Actual numbers:
Buoyancy:
For a given launch energy $T_0$, an object with a 70 liter displacement and a mass slightly more than 70kg, buoyancy force B.
Let it weigh $70kg*gravities$ on an extremely cold day. Then it will weigh $0.02 kilogram*gravities$ more on an extremely hot day than on an extremely cold day, because the density of air is about $0.3 kg/m^3 more at very low outdoor temperatures than at very high outdoor temperatures, holding pressure and humidity constant.
Object will attain a maximum height at $(mg+mB)\Delta h=T_0$, so for $mg+mB=70.02kg *g$, the change in launch height due to a 30% buoyancy decrease is about 0.03% lower on the hot day than on the cold day.
Air resistance (back-of-envelope approximation):
Let $g = 10m/s^2$.
Let the cross-sectional area be $A=0.1m^2$
let the drag coefficient be $C_d=2$
let the jump height be $h=0.8m$
Then the jumper's velocity was
$v = g \sqrt{h / (0.5g)} - gt =a - gt$
where a = 4m/s
over the domain 0<t<0.4s
and $\langle v^2 \rangle = (\int v^2 dt) /0.4s = 5.3m^2/s^2$
Providing the drag equation works across the temperature gap and the drag coefficient can be approximated as a constant for the launched object (preferably not a human)
Then $\langle F_d \rangle=0.5 \rho C_d A \langle v^2 \rangle = 0.528\rho * m^4/s^2$
Acting over 0.8m that gives us work of $W_d=0.4224 \rho *m^5/s^2$
Density of air is about 1 kg per cubic meter, so if $\Delta \rho = 0.3 \rho_0, \Delta W_d = 0.12672J \approx 0.1J$
Energy of a 0.8m launch height for a 70kg mass is in 560J, so the change in launch height due to a 30% air resistance decrease is about 0.02% higher on the hot day than on the cold day.
