It's somewhat hard to describe this with text. But if water were an sp3 hybrid, the two non-bonding lone pairs would be equal in energy. That would lead us to expect two peaks in water's XPS spectrum; one from the two equivalent hydrogen 1s-oxygen sp3 sigma bonds, and one from the two equivalent oxygen sp3 lone pairs. In fact, the XPS spectrum shows three peaks; one for two equivalent hydrogen 1s-oxygen 2p sigma bonds, one from a non-bonding oxygen 3p orbital (one of the lone pairs) and one from the non-bonding oxygen 2s orbital (the other lone pair) that's almost 5 eV lower in energy than the peak from the sigma bonding orbitals. This indicates that the 2s orbital on the oxygen is not actually hybridized with the 2p orbitals. The fact that you get approximately the correct geometry for the molecule if you consider it an sp3 hybrid is just a coincidence. Edit: If you want to really discuss the electronic structure and bond nature in water you would need to discuss its molecular orbital diagram, but the MO diagram just moves it even further from being an sp3 hybrid.