1938 U.S. National Championships – Women's singles

Second-seeded Alice Marble defeated Nancye Wynne 6–0, 6–3 in the final to win the women's singles tennis title at the 1938 U.S. National Championships.^[1][2][3]

Women's singles
1938 U.S. National Championships
Champion	Alice Marble
Runner-up	Nancye Wynne[1]
Score	6–0, 6–3
Draw	64
Seeds	16

Seeds

The tournament used two lists of eight players for seeding the women's singles event; one for U.S. players and one for foreign players. Alice Marble is the champion; others show in brackets the round in which they were eliminated.^[4]

Helen Jacobs (third round) Alice Marble (champion) Dorothy Bundy (semifinals) Sarah Fabyan (semifinals) Gracyn Wheeler (third round) Barbara Winslow (third round) Helen Pedersen (third round) Dorothy Workman (third round)

Jadwiga Jędrzejowska (quarterfinals) Simonne Mathieu (quarterfinals) Kay Stammers (quarterfinals) Nancye Wynne (finalist) Thelma Coyne (third round) Margot Lumb (quarterfinals) Nell Hopman (third round) Dorothy Stevenson (third round)

Draw

Final eight

	Quarterfinals							Semifinals							Finals
	(6)	Margot Lumb	4	7	1
	(4)	Nancye Wynne	6	5	6
								(4)	Nancye Wynne	5	6	8
								3	Dorothy Bundy	7	4	6
	3	Dorothy Bundy	6	3	6
	(2)	Simonne Mathieu	3	6	0
															(4)	Nancye Wynne	0	3
															2	Alice Marble	6	6
	2	Alice Marble	6	6	6
	(3)	Kay Stammers	8	3	0
								2	Alice Marble	5	7	7
								4	Sarah Fabyan	7	5	5
	4	Sarah Fabyan	6	6
	(1)	Jadwiga Jędrzejowska	1	4