Jamie Carragher states his prediction for Liverpool FC v Everton
The Liverpool FC legend is backing Jurgen Klopp's side to beat Everton in the Merseyside derby on Sunday afternoon
Jamie Carragher is backing Liverpool FC to beat Everton in the Merseyside derby at Anfield on Sunday afternoon.
The Reds will welcome Sam Allardyce’s side to Anfield in the former England manager’s first taste of the Merseyside derby since his appointment Everton boss.
Liverpool FC have been in top form over the past month or so following a six-game unbeaten run in the Premier League to climb up the top-flight table.
Jurgen Klopp’s side were 5-1 winners against Brighton and Hove Albion in the Premier League last weekend thanks to goals from Emre Can, Philippe Coutinho and Roberto Firmino.
Liverpool FC eased to a resounding 7-0 victory over Spartak Moscow in their final Champions League group stage fixture to secure their place in the last-16 clash.
Retired Liverpool FC defender Carragher is tipping the Reds to ease to a comfortable victory over Everton in the Merseyside derby at Anfield on Sunday afternoon.
“Is there a feel-good factor going into it after what happened in midweek? Any player or manager coming into this city knows how important this game is,” Carragher told Sky Sports News.
“As a fan, you are a little bit nervous thinking Sam Allardyce has come in and Everton have picked up a little bit and could it change?
“But then you come here in midweek and watch how well Liverpool are playing and it has to give you confidence.
“And with Liverpool being at home, you have to be confident Liverpool will get the three points. But in a derby game you always fear the worst…”
Liverpool FC finished in fourth place in the Premier League table to secure a return to the Champions League after a two-season absence.
The 18-time English champions signed four players in the summer transfer window to bolster their squad for a return to Europe’s premier club competition.
The Reds haven’t lost a Premier League game since a 4-1 loss at Tottenham Hotspur.