Question

# Rose had ten cats and fifteen dogs and was interested in knowing whether the cats and...

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?

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:
p1 = 6/10 = 0.6,
p2 = 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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