## 31.8 Example: Pet birds

A study examined people with lung cancer, and a matched set of similar controls who did not have lung cancer, and compared the proportion in each group that had pet birds (Kohlmeier et al. 1992).

These data were studied in Sect. 25.6; the data are shown again in Table 31.4, and the numerical summary in Table 31.5 (the computations are shown in Sect. 25.6).

Adults with lung cancer | Adults without lung cancer | Total | |
---|---|---|---|

Kept pet birds | 98 | 101 | 199 |

Did not keep pet birds | 141 | 328 | 469 |

Total | 239 | 429 | 668 |

One RQ in the study was:

Are the

oddsof having a pet bird the same for peoplewithlung cancer (cases) and for peoplewithoutlung cancer (controls)?

The parameter is the population OR, comparing the odds of keeping a pet bird, for adults with lung cancer to adults who do not have lung cancer.

The RQ could also be written as:

- Is the
**percentage**of people having a pet bird the same for people*with*lung cancer (cases) and for people*without*lung cancer (controls)? - Is the
**odds ratio**of people having a pet bird, comparing people*with*lung cancer (cases) and for people*without*lung cancer (controls), equal to one? - Is there a
**relationship**between having a pet bird and having lung cancer?

From this RQ (which is written in terms of *odds*),
the hypotheses could be written as:

- \(H_0\): The
*odds*of having a pet bird is*the same*for people*with*lung cancer (cases) and for people*without*lung cancer (controls). - \(H_1\): The
*odds*of having a pet bird is*not the same*for people*with*lung cancer (cases) and for people*without*lung cancer (controls).

The null hypothesis could also be written as:

- The
**percentage**of people having a pet bird is*the same*for people*with*lung cancer (cases) and for people*without*lung cancer (controls). - The
**odds ratio**of people having a pet bird, comparing people*with*lung cancer (cases) and for people*without*lung cancer (controls), is equal to one. - There is
**no relationship**between having a pet bird and having lung cancer.

Begin by **assuming** the null hypothesis is true:
no difference exists between the odds in the *population*.
Based on this assumption,
the **expected** counts can be found.

From the data
(Table 31.4),
overall
\(199\div 668 = 29.79\)% of people own a pet bird.
If there really was no difference in the odds (or the percentages)
of owning a pet bird between those with and without lung cancer,
about \(29.79\)% of the people in *both* lung cancer groups
are **expected** to own a pet bird.

Odds of keeping pet bird | Percentage keeping pet bird | Sample size | |
---|---|---|---|

With lung cancer: | 0.6950 | 41.0% | 239 |

Without lung cancer: | 0.3079 | 25.5% | 429 |

Odds ratio: | 2.26 |

About 29.79% of the 239 lung-cancer cases (or 71.20)
would be expected to have a pet bird,
and
about 29.79% of the 429 non-lung-cancer cases (or 127.80)
would be expected to have a pet bird.
A table of these *expected counts*
(Table 31.6).
shows that all expected counts are greater than five.
In practice,
you do not need to compute the expecte counts;
software does this automatically.

Adults with lung cancer | Adults without lung cancer | Total | |
---|---|---|---|

Kept pet birds | 71.2 | 127.8 | 199 |

Did not keep pet birds | 167.8 | 301.2 | 469 |

Total | 239.0 | 429.0 | 668 |

The numbers in Table 31.6
are what is *expected*,
*if* the percentage of people owning a pet bird
is the same for lung cancer and non-lung cancer cases.
How close are the expected and observed counts
(in Table 31.4)?

To compare the sample statistic
(what we **observed**) with the
hypothesised population parameter,
software is used to compute the value of \(\chi^2\)
(jamovi: Fig. 31.4;
SPSS: Fig. 31.5):
\(\chi^2=22.374\),
approximately equivalent to a \(z\)-score of

\[ \sqrt{22.374/1} = 4.730, \] which is very large. Hence, a small \(P\)-value is expected.

The software shows that the \(P\)-value
is very small
(\(P<0.001\)).
As usual,
a small \(P\)-value means that there is
very strong evidence
supporting \(H_1\), if \(H_0\) is assumed true.
That is,
the evidence suggests there is a *difference*
in the odds in the *population*.
We write:

The

sampleprovides very strong evidence (\(\chi^2=22.374\); two-tailed \(P<0.001\)) that the odds in thepopulationof having a pet bird is not the same for people with lung cancer (odds: 0.695) and for people without lung cancer (odds: 0.308; OR: \(2.26\); 95% CI from \(1.6\) to \(3.2\)).

**Think 31.2 (Interpretation)**This doesn’t imply that owning a pet bird

*causes*lung cancer. Why not?

*observational*. Confounders may explain the relationship (can you think of one?). In addition, may be having lung cancer means that people seek companionship in the form of a pet.