My wife is a saint. She knows that I am a proud man who often refuses to admit w

Question

My wife is a saint. She knows that I am a proud man who often refuses to admit when I am wrong. But she also knows that I take statistics very seriously, and that I will listen to compelling numbers. One day, early in our marriage, after I had lost yet another Scrabble game, I said, “You know, I think you and I are pretty equally matched!” She laughed (more than she should have) and pointed out that she’d won 5 out of the last 8 games we’d played. Deep down inside I knew she was right, but the numbers weren’t on her side…at least not yet. I pointed out that winning 5 out of 8 games is not very convincing evidence that she’s better than me. In fact, we’d need quite a bit more data before she could claim that she really was better. How much more data? Suppose that Melissa continues to win 62.5% (5 out of every 8) of the games of Scrabble we play. How many games will we need to complete before I must accept the fact that she really is a better player than I am?

This quesiton is based on a coin flipping simulation, and its resulting histogram. I need to get a P value of less than 0.05. How many more games must they play in order for that to happen? And how do I get there?

Additionally, this is the coin flipping simulation we are using: https://k300apps.shinyapps.io/coin-sampling/

Explanation / Answer

Here as we first assume that both of melissa and her husband are equally good at scrabble.

here if n is the number of such required games to differentiate between these two players.

so here p^ = 0.625 = x/n where x is the number of games won by melissa

so here,

as we need P- value less than 0.05

Pr(p >= 0.625) <= 0.05

Here p^ = 0.5

se(p) = sqrt (0.5 * 0.5/n) = 0.5/ sqrt(n)

so Z = (0.625 - 0.5)/ (0.5/sqrt(n) = n /4

as for p - value 0.05, Z = 1.96

n/4 >= 1.96

n >= 1.96 * 4

n > = 61.465 or n = 62

we get this by assuming the normal approximation to binomial. If we dont approximate it, we will get it for n = 51

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