Question

Of the sample of 72 observations for sparrows that survived, the mean humerus length was 0.74 in, and the standard deviation was 0.02 in. And of the sample of 64 observations for sparrows that died, the mean humerus length was 0.73 in, and the standard deviation was 0.03 in. Let μs denote the mean for the population distribution of humerus length for sparrows that would survive, and let μd denote the mean for the population distribution of humerus length for sparrows that would not survive. Bumpus would probably have pointed out that sparrows that survived tended to have a larger humerus length than those that did not. While this is true for his samples, it is not clear if we can infer from this that this is true in general (i.e., μs>μd). Conduct a significance test to determine if the difference in the sample means is statistically significant or not. Also estimate the difference in mean humerus length of sparrows that would survive the storm (μs) and those that would not (μd) using a confidence interval.

Answer 1

a)

Ho: there is no significant difference in the mean humerus length between sparrows that survived and sparrows that died. Us = ud

H1: sparrows that survived tended to have a larger humerus length than those that did not. Us > ud

	Group 1	Group 2
n=	72	64	COUNT
mean=	0.74	0.73	AVERAGE
s=	0.0200	0.0300	STDEV
s^2/n	0	0

Sp^2
{(n1-1)*s1^2 + (n2-1)*s2^2}/{n1+n2-2}
=((72-1)*0.02^2+(64-1)*0.03^2)/(72+64-2)
0.0006

t=
(0.74-0.73)/SQRT(0*(1/72+1/64))
2.3098

df= n1+n2-2
72+64-2
134

Alpha= 0.05

t(a, df)
=t(0.05,134)
=abs(t.INV(0.05,134)) =
1.656

p-value
P(T>t)
(1-P(T<2.3098)
T.DIST.RT(2.3098,134)
0.011213933

With (t=2.3098, p<5%),I reject the null hypothesis at 5% level of significance. And conclude that sparrows that survived tended to have a larger humerus length than those that did not. Us > ud

b)

CI = (Xbar1-Xbar2) +-t(a/2,df)*sqrt(Sp^2*(1/n1+1/n2))
lower =(0.74-0.73)-1.97782575808712*SQRT(0.0006*(1/72+1/64))= 0.0017
upper =(0.74-0.73)+1.97782575808712*SQRT(0.0006*(1/72+1/64))= 0.0183

I am 95% confidence that estimated population difference in mean humerus length of sparrows that would survive the storm (μs) and those that would not (μd) lie in the interval (0.0017, 0.0183). Since this confidence interval does not contain 0, there is sufficient evidence to conclude that at sparrows that survived tended to have a larger humerus length than those that did not.