0
$\begingroup$

Manufacturer of pharmaceutical products has to decide about the recovery from a certain disease for a new medication on basis of samples. For the test $H_0 : θ_0 ≥ 0.90$ versus $H_1 : θ_0 < 0:90$, we can only calculate type II error probabilities for specific values for $θ_1$ in $H_1$. Suppose that the manufacturer has in mind a specific alternative, say $θ_1 = 0.60$. His test statistic is x, observed number of successes in $n = 20$ trials and he will accept $H_0$ if x ≥ 15. Evaluate probabilities α and β.

SOLUTION

Acceptance region for $H_0$ is given by {$15; 16; 17; 18; 19; 20$}.

$$α = \text{P(type I error)} = P(x < 15|θ = 0.90) = 0.0114$$ $$β = \text{P(type II error)} = P(x ≥ 15|θ = 0.60) = 0.1255$$

β can be reduced by appropriately changing critical region: If critical region is $x < 16$ we get $α = 0.0433$ and $β = 0.0509$

What is the exact calculation behind $α = P(x < 15|θ = 0.90) $ and $β = P(x ≥ 15|θ = 0.60)$ ? I am aware we are dealing with binomial distribution from a small sample, so approximation to normal is not appropriate.

$\endgroup$
1
  • 1
    $\begingroup$ Yes, approximation is not appropriate. There are tables containing values of the cdf of a Binomial random variable, I suppose the parameter values $n=20$ and $p=0.9$ or $p=0.6$ can be found in many of them. However, nowadays you would just ask statistical software for the precise values. In R for instance, the commands would be "pbinom(14,20,0.9)" and "1-pbinom(14,20,0.6)". The exact calculation is of course based on $P(X=x|n,\theta)={n \choose x}\theta^x (1-\theta)^{n-x}$ $\endgroup$ – Mau314 Aug 15 '18 at 9:21
3
$\begingroup$

To compute the exact binomial probabilities in this problem, you could use (i) the binomial PDF formula, (ii) a statistical calculator programmed with the binomial PDF and CDF, or (iii) statistical software on a computer.

I will illustrate the use of R statistical software, and indicate how to use the binomial PDF.

Type I error. Suppose $X \sim \mathsf{Binom}(n=20, p=.9).$ Then $\alpha = P(X < 15) = P(X \le 14) = 0.0113.$

In R statistical software pbinom is a binomial CDF.

pbinom(14, 20, .9)
[1] 0.01125313

The same result can be obtained by adding terms of the binomial PDF function dbinom; the notation 0:14 is shorthand for a list of the numbers $k = 0, \dots, 14.$ These are the numbers in the 'Rejection region' of your test.

sum(dbinom(0:14, 20, .9))
[1] 0.01125313

By either computation, this differs slightly from the answer given in your Question, and the first computation in R agrees with @Mau314's Comment.

Using the binomial PDF would require summing terms $P(X = k) = {20 \choose k}(.9)^k(.1)^{n-k},$ for $k = 15, \dots, 20,$ and subtracting the sum from $1.$

Type II error. Suppose $X \sim \mathsf{Binom}(n=20, p=.6).$ Then $\beta(p=.6) = P(X \ge 15) = 1 - P(X \le 14) = 0.1256.$

1 - pbinom(14, 20, .6)
[1] 0.125599

Using the binomial PDF would require summing terms $P(X = k) = {20 \choose k}(.6)^k(.4)^{n-k},$ for $k = 15, \dots, 20.$

The 'power' of the test against the specific alternative $p = .6$ is $\pi(p=.6) = 1 - \beta(p=.6) = 1 - 0.1256 = 0.8744.$


The figure below shows the PDFs of the two binomial distributions used above. The Rejection region is to the left of the vertical broken line and the Acceptance region is to the right.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.