I see, so you entered 0.7/0.3 as p to account for
Post# of 148187
I just think one needs to use a two independent proportions test, not a simple binomial...
https://www.statsdirect.com/help/proportions/unpaired.htm
https://www.google.com/url?sa=t&source=we..._hbSwf8eKU
Will need to wait for others to see which calculator is correct...