Question

Functionally important traits in animals tend to vary little from one individual to the next within populations, possibly because individuals who deviate too much from the mean have lower fitness. If this is the case, 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 three-toed sloth (Bradypus variegatus). The authors proposed that since sloths don't move their heads much, the functional importance of SC is reduced, and may vary more than it does in more active animals. They obtained the following measurements of the ratio of the length to width of the anterior SC in 7 sloths. Assume this represents a random sample. In other, more active animals, the standard deviation of this ratio is 0.09.
Sloth CW Ratios
1.5
1.09
0.98
1.42
1.49
1.25
1.18
Fill in the blank for a with the estimate of the standard deviation of this measurement in three-toed sloths to two decimals, and include the leading zero
The 95% confidence interval for the standard deviation of this data is < σ < (two decimals - include the leading zero)
Does this interval include the value obtained from other species? (answer yes or no in blank d)

Answer 1

Answer:

Step-by-step explanation:

Hello!

Given the variable

X:  Ratio of the length to width of the anterior semicircular canals (SC) of the three-toed sloth.

The researcher's claim is that the ratio of the SC of the sloths is more variable than in other animal species that are more active.

For more active species the standard deviation of the ratio is σ= 0.09

1)

To calculate the sample standard deviation you have to calculate the sample variance first:

$S^2= \frac{1}{n-1} [sumX^2-\frac{(sumX)^2}{n} ]$

n=7; ∑X= 8.91; ∑X²= 11.8599

$S^2= \frac{1}{6} [11.5899-\frac{(8.91)^2}{7} ]= 0.0287= 0.029$

S= √S²= √0.029= 0.169≅ 0.17

The sample standard deviation of the ratio is 0.17

2)

The parameter of interest is the population standard deviation. To calculate a confidence interval for the standard deviation of a population you have to estimate the population variance first. Then calculate the square root of both limits of the interval for the variance to obtain the interval for the standard deviation.

The statistic to use is the Chi-Square and the formula for the interval is:

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

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

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

$[\frac{6*0.029}{14.449} ;\frac{6*0.029}{1.2373} ]$

[0.0120; 0.1406]

Using a 95% confidence level you'd expect the interval [0.0120; 0.1406] to include the true value of the population variance of the ratio of the length to width of the anterior semicircular canals (SC) of the three-toed sloths.

Now you have to calculate the square root of each limit:

[√0.0120; √0.1406]

[0.1097; 0.3750]

Using a 95% confidence level you'd expect the interval [0.1097; 0.3750] to include the true value of the population standard deviation of the ratio of the length to width of the anterior semicircular canals (SC) of the three-toed sloths.

3)

As you can see the calculated interval doesn't include the value obtained for the other species.

I hope this helps!