I am trying to test whether the sex ratio of some sampled individuals significantly differs from the expected sex ratio of 1. I have n= 64, of which female=34 and male=30.

I ran a binomial test:

`succ <- c(34,30) binom.test(succ,0.5) data: succ number of successes = 34, number of trials = 64, p-value = 0.708 alternative hypothesis: true probability of success is not equal to 0.5 95 percent confidence interval: 0.4023098 0.6572035 sample estimates: probability of success 0.53125 `

I would like to calculate the statistical power of this test, and I know that *power = 1-β*, where β is the type II error.

I am getting confused when reading this explanation. I don't understand how to adapt this formula (for different choices of n) to my case:

`enn = 1:2000 critical = qbinom(.025, enn, .5) beta = pbinom(enn-critical,enn,.55) - pbinom(critical-1,enn,.55) `

What I did was

`1-(pbinom(34,64,0.5)- pbinom(30, 64, .5)) [1] 0.7410237 `

but I am not sure if it is correct to use 0.5 as probability.

Moreover, I tried a different method, and I get a completely different result

`pwr.p.test(ES.h(.53125,.5),n=64, power=NULL, alternative = "two.sided") proportion power calculation for binomial distribution (arcsine transformation) h = 0.06254076 n = 64 sig.level = 0.05 power = 0.07913605 alternative = two.sided `

Is one of these two tests correct and why?

Thanks for your help!

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#### Best Answer

In order to find 'power', you need to have a specific alternative in mind. Suppose your null hypothesis is $H_0: p = 0.5$ vs. $H_a: p > 0.5,$ where $p = P(mathrm{Female}).$ Also suppose you have $n = 64$ and you want the power of a test at level $alpha = 0.05$ against the specific alternative $p = 0.6.$

For an exact binomial test, you need to find the critical value $c$ such that $P(X ge c,|,n=64, p=.5)$ is maximized, but still below $0.05.$ In R, where `dbinom`

, `pbinom`

, and `qbinom`

denote binomial PDF, CDF, and quantile function (inverse CDF), respectively, we see that the critical value is $c = 40.$ Notice that, because of the discreteness of binomial distributions, the so-called `5% level' actually rejects with probability $P(mathrm{Rej}, H_0 | H_0, mathrm{True}) approx 3%.$

`qbinom(.95, 64, .5) [1] 39 sum(dbinom(39:64, 64, .5)) [1] 0.05171094 sum(dbinom(40:64, 64, .5)) [1] 0.02997059 1 - pbinom(39, 64, .5) [1] 0.02997059 `

Then the power of this test against alternative value $p = 0.6$ is given by $P(X ge 40,|,n=64, p=0.6) = 0.3927.$

`1 - pbinom(39, 64, .6) [1] 0.392654 `

We can make a 'power curve' for this test by looking at a sequence of alternative values `p.a`

between $0.5$ and $.75.$ The first block of R code below makes the solid black line in the plot below.

`p.a = seq(.50, .75, by=.01) p.rej = 1 - pbinom(39, 64, p.a) plot(p.a, p.rej, type="l", main="Power Curve") abline(h=c(.03,1), col="green2") `

If we look at a level $alpha = 0.05$ test of $H_0: p = 0.5$ vs $H_a: p > 0.5$ with $n = 256$ subjects, then the critical value is $c = 141,$ the rejection probability when $H_0$ is true is $0.046,$ and the power against various alternative values of $p$ is greater, as shown by the dotted blue line in the figure.

`c.256 = qbinom(.95, 256, .5); c.256 [1] 141 1 - pbinom(c.256, 256, .5) [1] 0.04565604 p.rej.256 = 1 - pbinom(c.256, 256, p.a) lines(p.a, p.rej.256, col="blue", lty="dotted") `

*Notes:* Because $n = 64$ is sufficiently large to use normal approximations, you might want to try using normal approximations. A disadvantage is that this ignores the issue of discreteness, so it may appear that your test rejects exactly 5% of the time when $H_0$ is true. Also, you'd need to use a continuity correction for best results.

One relevant computation for the significance level in R is:

`1 - pnorm(39.5, 32, 4) [1] 0.03039636 `

(Approximate) power is $0.3895:$

`mu.a = 64*.6; sg.a = sqrt(64*.6*.4) mu.a; sg.a [1] 38.4 [1] 3.919184 1 - pnorm(39.5, mu.a, sg.a) # Using NORM(mu.a, sg.a) [1] 0.3894815 1 - pnorm((39.5 - mu.a)/sg.a) # Standardizing and using NORM(0,1). [1] 0.3894815 `

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