Weddell seals live in the Antarctic and feed on fish during

Weddell seals live in the Antarctic and feed on fish during long, deep dives in freezing water. The seals benefit from these feeding dives, but the food they gain comes at a metabolic cost. The dives are strenuous. A set of researchers wanted to know whether feeding per se was also energetically expensive, over and above the exertion of a regular dive (Williams et al. 2004). They determined that the metabolic cost of dives by measuring the oxygen use of seals as they surfaced for air after a dive. They measured the metabolic cost of 10 feeding dives and for each of these also measured a nonfeeding dive by the same animal that lasted the same amount of time. The data, in (ml O2 kg^-1), are as follows: a. Estimate the mean change in oxygen consumption during feeding dives compared with nonfeeding dives. b. What is the 99% confidence interval for the population mean change? c. Test the hypothesis that feeding does not change the metabolic costs of a dive.

Solution

(a)

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(b)

Given a=1-0.99=0.01, t(0.005, df=9) =3.25 (from student t table)

So the lower bound is

mean different - t*s/vn = -31.78 - 3.25*7.2962/sqrt(10) =-39.2786

So the upper bound is

mean different + t*s/vn = -31.78 + 3.25*7.2962/sqrt(10) =-24.2814

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(c) Let mu_d be the mean difference between nonfeeding and feeding

The test hypothesis:

Ho: mu_d =0 (i.e. null hypothesis)

Ha: mu_d not equal to 0 (i.e. alternative hypothesis)

The test statistic is

t= mean difference /(std. dev. /vn)

=-31.78/(7.2962/sqrt(10))

=-13.77

It is a two-tailed test.

Given a=1-0.99=0.01, the critical values are t(0.005, df=9) =3.25 or -3.25 (from student t table)

The rejection regions if t<-3.25 or t>3.25, we reject the null hypothesis.

Since t=-13.77 is less than -3.25, we reject the null hypothesis.

So we can conclude that feeding change the metabolic costs of a dive

74.6800	mean nonfeeding
106.4600	mean feeding
-31.7800	mean difference (nonfeeding - feeding)