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Suppose X is a random variable on {0, 1, 2, . . .} with unknown p.m.f. p(x). To test the hypothesis $H_{0}$ : X ∼ $Poisson(1/2)$ against $H_{1}$ : p(x) = $2^{−(x+1)}$ for all x ∈ {0, 1, 2, . . .}, we reject $H_{0}$ if x > 2. The probability of type-II error for this test is

(A) $1/4$

(B) $1−(13/8)e^{−1/2}$

(C) $1 −(3/2)e^{−1/2}$

(D) $7/8$

When I tried this question the answer I got was $(3/2)e^{-1/2}$ I checked and rechecked the question and Reread the definition of Type-II error but to no avail Can you tell me where I went wrong?

enter image description here

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  • $\begingroup$ Can you share your solution as well? $\endgroup$ – gunes Sep 18 at 6:45
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    $\begingroup$ OK.I DID(Do excuse the handwritingT^T) $\endgroup$ – Muskaan Madan Sep 18 at 6:57
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    $\begingroup$ Thanks for showing your work. // Probability of Type II error is computed using the distribution of the alternative hypothesis. $\endgroup$ – BruceET Sep 18 at 8:17
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    $\begingroup$ The answer you got explains that you used the wrong pdf. Aside from that, I find the use of 'type II error' in this exercise a bit confusing. In this example you compare $H_0$ versus a specific alternative $H_1$. That is fine and clear, but with the type II error is ambiguous. Personally I associate it more with the situation when $H_0$ is not true rather than when $H_1$ is true; There are more ways (than $H_1$) how $H_0$ can be not true. So the use of the term 'type II error' is in a narrow sense. I would prefer to use the term power instead. $\endgroup$ – Sextus Empiricus Sep 18 at 8:23
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    $\begingroup$ Continued... The type II error is sometimes computed by integrating over all the alternatives $H_a$ and some known or estimated probability that $H_a$ may occur in practice. This hypothetical dichotomous setting where the only alternative to $H_0$ is a single $H_1$, that is a bit weird. (It may actually occur in practice where you have classes, e.g. a person is 'sick' or or 'not-sick', but even there you might consider that reality is more nuanced and you have people of different degrees of sickness) $\endgroup$ – Sextus Empiricus Sep 18 at 8:33
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What you are interested in is the probability of $p(x)=2^{-(x+1)} \leq 2$.

> sum(2^-(0:2+1))
[1] 0.875

which is answer (D).

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    $\begingroup$ And so the mistake that the OP made was using the wrong pdf. $\endgroup$ – Sextus Empiricus Sep 18 at 7:42
  • $\begingroup$ I was interrupted getting my Answer finished. Saw you answer (+1) when I posted mine. $\endgroup$ – BruceET Sep 18 at 8:10
  • $\begingroup$ Lol. I didn't even realize that 😅😅 $\endgroup$ – Muskaan Madan Sep 18 at 9:42
  • $\begingroup$ Thanks for your help $\endgroup$ – Muskaan Madan Sep 18 at 9:42
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Your rejection region is $\{X < 2\},$ so the significance level is $\alpha = 1- P(X \le 2),$ where $X \sim \mathsf{Pois}(1/2).$ As computed in R, $\alpha \approx 0.0144.$

1 - ppois(2, 1/2)
[1] 0.01438768

Let's begin by looking at a a graph of the probability distributions according to $H_0$ (blue) and $H_a$ (brown).

x = 0:15
pdf.0 = dpois(x, 1.2)
pdf.a = 2^(-(x+1))
hdr = "Null (blue) and Alternative Dist'ns"
plot(x-.1, pdf.0, type = "h", ylim=c(0,.5), col="blue", 
     lwd=2, ylab="PDF", xlab="x", main=hdr) 
 lines(x+.1, pdf.a, type="h", col="brown", lwd=2)
 abline(v = 2.5, col="red", lty="dotted")

enter image description here

The probability of Type II for this test will not be small because the distributions under $H_0$ and $H_a$ are so nearly alike. (This may be what put you on a wrong path in your attempt to answer. In practice, useful tests tend to be ones for which Type I and Type II errors are both relatively small.)

The probability of Type I Error (significance level) $\alpha$ is the sum of the heights of the blue bars in the rejection region (to the right of the vertical dotted line).

The probability of Type II Error is the probability of failing to reject the null hypothesis when it is false. So the probability of Type II error is the sum of the heights of the brown bars to the left of the vertical dotted line: $1/2 + 1/4 + 1/8 = 7/8.$ Answer (D).

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  • $\begingroup$ Don't know the way around the graphs, so it's a bit tough for me. But thanks anyway $\endgroup$ – Muskaan Madan Sep 18 at 9:42
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    $\begingroup$ It's worthwhile figuring how to make a graph of a the PDF of a distribution. Not complicated. Just write down the values the random variable can take along with the probabilities. (in this case there are infinitely many values, so you stop your list when the probabilities get to small too matter.) Then make the plot. // There must be a few such plots in examples in your textbook. // Maybe ask your instructor about making graphs of PDFs during class or in office hours. (If this is a class hastily converted to distance learning because of C-19, your instructor may appreciate the interaction.) $\endgroup$ – BruceET Sep 18 at 17:04

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