The relationship among pressure, density, and temperature is not simple. It is described by the ideal gas law:
$$ P=\frac{\rho k T}{\bar{m}}$$
where $P$ is the pressure, $\rho$ is the density, $T$ is the absolute temperature (in Kelvin or Rankine, not degrees Celsius or Fahrenheit), $k$ is the Boltzmann constant ($1.38\times 10^{-23}~\rm J/K$) and $\bar{m}$ is the average mass of a molecule of air (approximately $4.8\times10^{-23}~\rm g$ for dry air).
Notably, this contains pressure, density, and temperature. So it's not as simple as "if temperature goes up, pressure goes up," since it is equally possible that "temperature goes up, pressure stays the same, so density must go down."
By considering the forces on a parcel of air along with the ideal gas law, and assuming the air is not moving, we can find a pressure lapse rate $dP/dz$ that turns out to be given by:
$$ \frac{1}{P}\frac{dP}{dz} = -\frac{\bar{m}g}{kT}$$
where $g$ is the acceleration of gravity, approximately $9.8~\rm m/s^2$.
Notice that as temperature increases, $dP/dz$ decreases. In other words, the pressure changes less quickly with altitude. This means even if the sea level pressure is standard, the pressure at 1000 feet will be higher than usual. Or, in other words, the pressure levels 29.92 (corresponding to sea level) and 28.86 (corresponding to a pressure altitude of 1000 feet) will be farther apart. So if you climb to 1000 feet of true altitude, you haven't yet reached that level, and your altimeter will indicate lower than 1000 feet.