Question

Rose had ten cats and fifteen dogs and was interested in knowing whether the cats and dogs differed in mouse-catching ability. She arranged a test and scored each animal as being “good” or “poor” as a mouse-catcher. Six of the cats and four of the dogs were scored as “good”. Is mouse-catching ability conditional on being a cat or dog?

Answer #1

The pooled proportion here is computed as:

P = (x1 + x2) / (n1 + n2) = (6 + 4)/(10 + 15) = 0.4

The standard error now is computed here as:

The sample proportions here are computed as:

p_{1} = 6/10 = 0.6,

p_{2} = 4/15 = 0.2667

The test statistic for the difference in proportions here is computed as:

As this is a two tailed test, the p-value here is computed from
the standard normal tables here as:

p = 2P(Z > 1.6667) = 2*0.0478 = 0.0956

As the p-value here is 0.0956 > 0.05 which is the level of
significance, therefore the test is not significant here and we
cannot reject the null hypothesis here. **Therefore
mouse-catching ability is conditional on being a cat or
dog**

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