1
$\begingroup$

enter image description here

website link

Here are some of the hypothesis statements, I don't know why do cases (i),(j) null hypothesis would not be another way around, and they also contradict to the case (b),(f)

Any ideas on how to correctly states the null and alternative hypotheses? Appreciate for any comments

$\endgroup$
1
  • 1
    $\begingroup$ Answers to parts (i) and (j) are OK. See my Answer for a discussion of (i). For a more general discussion, see this page, linked in the margin under 'Related'. $\endgroup$
    – BruceET
    Aug 18, 2021 at 9:41

1 Answer 1

1
$\begingroup$

(i) If you take the alternative hypothesis to be $H_a: p < 0.11,$ and you reject $H_0: p = 0.11$ or $H_0: p \ge 0.11$ at the 5% level, then it is because your sample proportion $\hat p$ of women developing breast cancer is sufficiently below $0.11$ to be called 'significantly' below $0.11$ at the 5% level. This would support the claim made in the text version of part (i).

Example: Perhaps you have $n = 1000$ women in your sample, of whom 87 developed breast cancer. Then the null distribution is $\mathsf{Binom}(n=1000,p=0.11)$ and $\hat p = 87/1000 = 0.087 < 0.11.$

Moreover, the P-value of an exact binomial test would be $0.01977 < 0.05 = 5\%$ so we could reject $H_0$ at the 5% level and say that $\hat p$ is significantly below $0.11$ at the 5% level of significance. (In fact, the P-value is sufficiently small to claim significance at the 1% level. [Computations in R.]

pbinom(87, 1000, .11)  # binomial CDF
[1] 0.009771923

Roughly speaking, you might say that the 'critical value' for this left-sided test at (about) the 5% level is $c=93,$ so that any number of breast cancer cases 83 or fewer out of the 1000 subjects would lead to rejection on $H_0$ at the 5% level (and any number of cases 87 or below would lead to rejection at the 1% level).

qbinom(.05, 1000, .11)  # quantile function
[1] 94
pbinom(94, 1000, .11)   # CDF
[1] 0.0562174
pbinom(93, 1000, .11)
[1] 0.04521503

qbinom(.01, 1000, .11)
[1] 88
pbinom(87, 1000, .11)
[1] 0.009771923

The figure below shows the null distribution with a vertical dotted line at the 5% critical value for a left=tailed test.

enter image description here

R code for figure:

x = 0:200;  PDF = dbinom(x, 1000, .11)
plot(x, PDF, type="h", col="blue", 
      main="Null Distribution")
abline(h=0, col="green2")
abline(v=0, col="green2")
abline(v=93.5, col="red", lwd=2, lty="dotted")
$\endgroup$
8
  • $\begingroup$ Thank you for your reply. I know how to interpret the hypotheses, but I don't know how to correctly states the null and alternative. For example, in example (b), the question specifies "......at most $60$% " so our null would be $p\leq 0.6$ and alternative is $p>0.6$.However in question (i) specifies "....under $11$%", why does null is not $p<0.11$ but $p\geq 0.11$ $\endgroup$
    – LJNG
    Aug 18, 2021 at 17:37
  • $\begingroup$ The interpretation and the formulation have to go together. Once you really know how to interpret $H_a$ in part (i), it should be clear that is is properly formulated. Think it through to the end; pretend you have data, consider why you are doing the project and how you hope the test of hypothesis turn out // In problems such as these, the null hypothesis is sometimes called the 'research' hypothesis'. The idea is that rejecting $H_0$ in favor of $H_a$ means your research project has met its goal. $\endgroup$
    – BruceET
    Aug 18, 2021 at 17:59
  • $\begingroup$ But be careful: in a goodness-of-fit test, you are sometimes hoping $H_0$ is true. (Genetic model seems "confirmed"; data seems normal, etc.) That may be one reason we get so many questions here about GOF testing. $\endgroup$
    – BruceET
    Aug 18, 2021 at 18:00
  • $\begingroup$ Thank you for your reply. I spent some time read over and over again of your reply, you explain the ideas behind the hypothesis, which is really good. But I still do not know how does it answer my question. According to your answer, I am not sure if I propose correctly. Based on the question, either way is correct! $H_0: P\geq 0.11$ and $H_a<0.11$ is same as $H_0: P\leq 0.11$ and $H_a>0.11$ former reject null, later do not reject null. $\endgroup$
    – LJNG
    Aug 23, 2021 at 13:11
  • 1
    $\begingroup$ (b) and (i) are for different situations. There is no reason they must agree. The prompt "less than 11%" is not mine, it's what's given for (i). // Notice that the null hypothesis always has and =-sign (whether (as $=,\le$ or $\ge.$ So your last sentence is not exactly true.// If you have one of these ten parts to discuss in more detail, please find an actual application that fits it and start a fresh Question. // Also please have a further look at the link in my Comment just below the Question. Not always obvious whether $H_a$ should have $<, >$ or $\ne.$ Requires practice. $\endgroup$
    – BruceET
    Aug 24, 2021 at 4:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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