Question

Functionally important traits in animals tend to vary little from one individual to the next within populations, possible because individuals that deviate too much from the mean die sooner or leave fewer offspring in the long run. If so, does variance in a trait rise after it becomes less functionally important? Billet et al. (2012) investigated this question with the semicircular canals (SC) of the inner ear of the three-toed sloth (Bradypus variegatus). Sloths move very slowly and infrequently, and the authors suggested that this behavior reduces the functional demands on the SC, which usually provide information on angular head movement to the brain. Indeed, the motion signal from the SC to the brain may be very weak in sloths as compared to faster-moving animals. The following numbers are measurements of the length to the width of the anterior semicircular canals in seven sloths. Assume that this represents a random sample.
1.52, 1.06, 0.93, 1.38, 1.47, 1.20, 1.16

a. In related, faster-moving animals, the standard deviation of the ratio of the length to the width of the anterior semicircular canals is known to be 0.09. What is the estimate of the standard deviation of this measurement in three toed sloths?
b. Based on these data, what is the most plausible range of values for the population standard deviation in the three-toed sloth? Does this range include the known value of the standard deviation in related, faster-moving species?
c. What additional assumption is required for your answer in (b)? What do you know about how sensitive the confidence interval calculation is when the assumption is not met?

Answer 1

Answer:

Step-by-step explanation:

Hello!

The objective is to study the semicircular canals (SC) of the inner ear of the three-toed sloth (Bradypus variegatus) to see if it is weak compared to faster animals.

The study variable is X: length to width of the anterior semicircular canal of a three-toed sloth.

A sample of 7 sloths was taken and the semicircular canal was measured:

1.52, 1.06, 0.93, 1.38, 1.47, 1.20, 1.16

∑X= 8.72

∑X²= 11.15

a.

$S^2= \frac{1}{n-1} * [sumX^2-\frac{(sumX)^2}{n} ]= \frac{1}{6} *[11.15-\frac{(8.72)^2}{7} ]$

S²= 0.047≅0.05

S=0.218≅ 0.22

Comparing the estimation of the variance of the length to width of the anterior semicircular canal of three-toed sloths with the known number of length to width of the anterior semicircular canal of faster animals (S=0.09), it appears that the variability os length to width of the anterior semicircular canal of sloths is greater than the length to width of the anterior semicircular canal of faster animals.

b. and c.

To estimate the most plausible range of values of the population standard deviation of the anterior semicircular canal of the sloths, you have to do an estimation per confidence interval.

To be able to make this estimation we have to assume that the variable of interest has a normal distribution. With this assumption, it is valid to use a Chi-Square statistic to estimate the population standard deviation.

$X^2= \frac{(n-1)S^2}{Sigma^2} ~X^2_{n-1}$

I'll choose a confidence level of 95%

The formula for the interval is:

$[\frac{(n-1)S^2}{X^2_{n-1;1-\alpha /2}} ;\frac{(n-1)S^2}{X^2_{n-1;\alpha /2}} ]$

$X^2_{n-1;1-\alpha /2}= X^2_{6;0.975}= 14.449$

$X^2_{n-1;\alpha /2}}= X^2_{6;0.025}= 1.2373$

$[\frac{6*0.05}{14.449} +\frac{6*0.05}{1.2373} ]$

[0.2076;2.4246] ⇒This confidence interval is for the population variance, calculating the square root of each bond gives us the CI for the population standard deviation:

√[0.2076;2.4246]= [0.1441;1.5571]

The 95% CI [0.1441;1.5571] is expected to contain the true value of the population standard deviation of the length to width of the anterior semicircular canal of the three-toed sloths.

As you can see this interval does not contain the known value of the population standard deviation for faster animals, which leads to thinking there is a difference between the standard deviation of the anterior semicircular canal in both species.

I hope it helps!