Question

A new species of sea crab has been discovered, and an experiment conducted to determine whether or not the animal can regulate its temperature. That the animal can maintain a body temperature different from the surroundings would be considered evidence of this regulating capability. Ten of these sea crabs were exposed to ambient temperatures of 24 degrees Celsius. Their body temperatures were measured with the results below:24.33, 24.61, 24.67, 24.64, 24.42, 24.97, 25.23, 24.73, 24.90, 24.44For purposes of this example, assume that it is reasonable to regard these 10 crabs as a random sample from the population of all crabs of this species.a)- Calculate a point estimate of the population mean.b)-Construct and interpret a 99% confidence interval for m.c)-Does it appear from these data that the crabs are able to regulate their body temperature? Provide statistical justification for your response.

Answer 1

Answer:

a) $\hat \mu =\bar X=24.694$

b) The 99% confidence interval would be given by (24.409;24.979)

c) Since the 99% confidence interval not contains the exposed to ambient temperatures of 24 degrees Celsius, and since the animals "that can maintain a body temperature different from the surroundings would be considered evidence of this regulating capability", on this case we can conclude that the sea crabs are able to regulate their body temperature.

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Data: 24.33, 24.61, 24.67, 24.64, 24.42, 24.97, 25.23, 24.73, 24.90, 24.44

$\bar X$ represent the sample mean  

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n=10 represent the sample size  

Part a

A point of estimate for the population mean is the sample mean, given by this formula:

$\bar X= \frac{\sum_{i=1}^n x_i}{n}$

If we apply this formula we got that $\hat \mu =\bar X=24.694$

Part b

In order to calculate the confidence interval first we need to calculate the sample deviation given by this formula:

$s=\frac{\sum_{i=1}^n (x_i -\bar x)}{n-1}$

If we use this formula we got that $s=0.277$

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)  

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=10-1=9$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,9)".And we see that $t_{\alpha/2}=3.25$

Now we have everything in order to replace into formula (1):

$24.694-3.25\frac{0.277}{\sqrt{10}}=24.409$    

$24.694+3.25\frac{0.277}{\sqrt{10}}=24.979$

So on this case the 99% confidence interval would be given by (24.409;24.979)    

Part c

Since the 99% confidence interval not contains the exposed to ambient temperatures of 24 degrees Celsius, and since the animals "that can maintain a body temperature different from the surroundings would be considered evidence of this regulating capability", on this case we can conclude that the sea crabs are able to regulate their body temperature.