Does this produce the anti-Markovnikov product because of the ROOR? I know that the anti-Markovnikov product is produced when there is a free radical present. Does the ROOR separate into two free radicals and that changes the reaction? But ROOR doesn't end up in the product at all right? Why? Is my thinking correct?
Yes, having a peroxide will change the regiochemistry to an anti-Markovnikov addition. The reason, as suggested, is because of a change in mechanism to a radical process (below). The homolytic decomposition of the peroxide initiates the radical chain, generating two hydroxyl radicals. The hydroxyl radical abstracts a hydrogen from HBr to give a bromine radical. This is where the cycle really begins, with bromine radical adding to the terminal end of the alkene, leaving a radical at the internal position. The stability of a tertiary radical relative to a primary radical drives this selectivity. The tertiary carbon radical abstracts a hydrogen from the next molecule of HBr, giving the product and a new bromine radical which reenters the radical chain.