#24 Chapter 12 Comparing two means 362 measure of the excitability of the males
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#24
Chapter 12 Comparing two means 362 measure of the excitability of the males in the two different circumstances, Six males were exposed to the scent of pre-ovulatory females their readings averaged 1.5I with a standard deviation of 0.25, Six different males exposed to post-ovulatory females averaged readings of 0.87 with a standard deviation of 0.31. Assume that the electro-olfactogram readings were approxi mately normally distributed within groups a. They determined the metabolic by measuring the oxygen use of seals surfaced for air after a dive Test for a difference in the excitability of the males with exposure to these two types of females. b. What is the estimated average difference in electro-olfactogram readings between the two groups? What is the 95% confidence abolic cost of 10 feeding dives of these also measured a nonfeeding r the same animal that lasted the sdiv and e me. The data, in (ml Ozkgt),are as fol means en consumption after nonfeeding 23. A baby dolphin is born into the ocean, which is a fairly cold environment. Water has a high heat conductivity, so the thermal regulation of a newborn dolphin is quite important. It has Individual been known for a long time that baby dolphins blubber is different in composition and quantity 59.8 from the blubber of adults. Does this make the babies better protected from the cold compared to adults? One measure of the effectiveness of blubber is its "conductance." This value was cal culated on six newborn dolphins and eight adult 32 96,1 81.3 81.3 dolphins (Dunkin et al. 2005). The newborn dol- phins had an average conductance of 10.44, with 104.1 a standard error of the mean equal to 0.69. The adult dolphins' conductance averaged 8.44, with the standard error of this estimate equal to 1.03 All measures are given in watts per square meter a. Estimate the mean change in oxygen co sumption during feeding dives compared with nonfeeding dives per degree Celsius b, what is the 99% confidence interval for Calculate the standard deviation of conduc- tance for each group population mean change? a. Test the hypothesis that feeding does n change the metabolic costs of a dive. c. b. Test the null hypothesis that adults and new- borns do not differ in the conductance of 25. Have you ever noticed that when you tear gernail, it tends to tear to the side and not into the finger? (Actually, the latter does bear too much thinking about.) Why mig be so? One possibility is that fingernails tougher in one direction than another. F al. (2004) compared the toughness of h fingernails along a transverse dimensio side) with toughness along a longitudin their blubber 24. 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).Explanation / Answer
Let, X1 and X2 be defined as,
X1: Oxygen consumption after non-feeding dive,
X2: Oxygen consumption after feeding dive.
Here, n=10, Mean(X1) = 74.68, Mean(X2) = 106.46, s(X1) = 19.393 and s(X2) = 24.0938.
(a) Mean change in oxygen consumption during feeding dives compared to non-feeding dives = Mean(X2) - Mean(X1) = 106.46 - 74.68 = 31.78.
(b) Suppose the population mean corresponding to X1 and X2 be mu1 and mu2.
99% confidence interval for (mu2 - mu1) is given by,
[ (Mean(X2) - Mean(X1)) - t0.005,9 sd/sqrt(10) , (Mean(X2) - Mean(X1)) - t0.005,9 sd/sqrt(10) ]
where sd = standard deviation of difference (X1 - X2) = 7.296 and t0.005,9 = 3.25.
So, the 99% CI is given by the interval [ 31.78 - 7.498, 31.78 - 7.498 ] = [ 24.282, 39.278 ].
(c) To test the required condition given in the question, we need to test whether the difference is significant or not. To test that hypothesis, the test statistic,
t = 31.78/(7.296/sqrt(10)) = 31.78 / 2.307 = 13.775 which is larger than t0.005,9 = 3.25. Hence, the null hypothesis ( that there is no difference in feeding) is rejected at 1% level of significance and we can conclude that there is significant differences between the two feeding. Hence, feeding change the metabolic cost of a dive.
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