Differences in weight

Question

George suspected that cats in Boise were heavier than cats in Pocatello. He went out, selected 10 cats at random in Boise and 10 cats at random in Pocatello, and weighed them. He found that the cats in Boise weighed on average 10 lb (SD = 2.1 lb), whereas the cats in Pocatello weighed 8 lb (SD = 1.8 lb). Can he conclude that cats in Boise differ from cats in Pocatello in terms of body weight? Why?

Answer 1

I am soo boored. Why else answer a 10 year old question? Well it is a classical two sample t-distribution hypothesis test so it is worth looking at.

Sample One: n₁ = 10 , x1_bar = 10lb s1 = 2.1lb

Sample Two: n₂ = 10 , x2_bar = 8lb s2 = 1.8lb

Since there is no reason to believe that there is a divergence in cats in such a boring place as Idaho we can used the pooled variance with degree of freedom = n₁+ n2 -2 or d.f. = 18

Pooled variance (s_p²) = ( (n₁-1)s₁² + (n₂-1)s₂² ) / (n₁+n₂-2). = (9*2.1² + 9*1.8²)/18 = 3.828

T_critical = (x_1bar - x_2bar) / √s_p²*(1/n₁ + 1/n₂) =(10-8) /√(3.828*.2) = 2.286

Using CDF for the T distribution and 18 d.f. the p-value (two sided) is .0346 or 3.46%

Can we answer the question? No, of course not because the question is, as is often the case, incomplete. Students fail in not knowing what critical information is needed to answer parametric statistics problems. In this case no alpha (α) or allowable type I error has been specified. Without it, we can't answer the problem.

One could make assumptions but assumptions could be wrong. Since a p value of .0346 is not very large compared to typical (but not mandatory) significance levels of .1, .05, or even .01 the answer is indeterminate. An assumption that we should reject the null hypothesis is probably not a bad assumption BUT THE QUESTION IS INCOMPLETE so there is no answer. Better luck next time. Ask the question properly!