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September 3

Sampling distribution of a difference between two independent sample proportions

Example 10-3 on slide 73 of Lloyd Jaisingh's Statistics for the Utterly Confused says "the probability that the difference in the proportion of success ... is at least 10 percent is 0.2877." (Direct link to slide 73) I think 10% never enters the question, and 0.2877 is the probability that the difference is at least 30% (from 0.7-0.4). Am I making sense? (Am I providing enough information?) BeforeBreakfast (talk) 01:08, 3 September 2011 (UTC)[reply]

It says "n₁=100, n₂=40, p₁=0.75, p₂=0.5, p̂₁=70/100=0.7, p̂₂=16/40=0.4. We need to determine P(p̂₁−p̂₂≥0.1). Substituting in the above equation to find the z score, we have that the z score value is 0.56. Thus, P(p̂₁−p̂₂≥0.1)=P(z≥0.56). ... That is, the probability that the difference in the proportion of success ... is at least 10 percent is 0.2877."
But I think we've computed P(p̂₁−p̂₂≥0.3), since 0.7−0.4=0.3. BeforeBreakfast (talk) 01:08, 3 September 2011 (UTC)[reply]

The equation being used is on slide 68. It's z=(p̂₁−p̂₂)−(p₁−p₂)/sqrt[(p₁(1−p₁)/n₁)+(p₂(1−p₂)/n₂)]. I filled this in as z=(0.7−0.4)−(0.75−0.5)/sqrt[(0.7(1−0.7)/100)+(0.4(1−0.4)/40)], so I agree z=0.56. I just think that z score corresponds to P(p̂1−p̂2≥0.3), not P(p̂1−p̂2≥0.1). I can't see where the 0.1 would come from at all. I hope I've explained well enough what I'm confused about. It seems to me that the text is wrong or I don't know what I'm doing. Any help is appreciated. BeforeBreakfast (talk) 01:33, 3 September 2011 (UTC)[reply]

I think you're right. It isn't 100% clear to me what the author intended, but the solution definitely seems wrong as written. Looie496 (talk) 02:47, 3 September 2011 (UTC)[reply]

Thanks, Looie. BeforeBreakfast (talk) 22:40, 3 September 2011 (UTC)[reply]