Question

Marine biologists have been studying the effects of acidification of the oceans on weights of male baluga whales in the Arctic Ocean. One of the studies involves a random sample of 16 baluga whales. The researchers want to create a 95% confidence interval to estimate the true mean weight of male baluga whales. Their data follow a normal distribution. The population standard deviation of weights of male baluga whales is ????=125 kg, and the researchers feel comfortable using this standard deviation for their confidence interval.
Assuming the relevant requirements are met, calculate the margin of error in estimating the true mean weight of male baluga whales in the Artic Ocean.

a. 15.31 kg
b. 51.40 kg
c. 61.25 kg
d. 80.49 kg

Assuming the relevant requirements are met, what sample size would be required if the researchers wanted the margin of error to be 45 kg?

a. 25
b. 30
c. 35
d. 40

Are the requirements for the use of a confidence interval met? Explain.

a. Yes. The distribution of sample means is normal because the data are normal.
b. Yes. The distribution of sample means is normal because the sample size is large.
c. No. The distribution of sample means is not normal because the sample size is small.
d. No. The fact that the data are normal does not imply that the distribution of sample means is normal.

Answer 1

Answer:

a) Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $t_{\alpha/2}=1.96$

And the margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$  

And replacing we got:

$ME=1.96\frac{125}{\sqrt{16}}=61.25$  

b) $ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$    (a)

And on this case we have that ME =45 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (b)

$n=(\frac{1.960(125)}{45})^2 =29.64 \approx 30$

So the answer for this case would be n=30 rounded up to the nearest integer

b. 30

c) And for this case the conditions in order to use the confidence interval are satisfied since:

a. Yes. The distribution of sample means is normal because the data are normal.

Step-by-step explanation:

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".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

$\sigma =125$ represent the population standard deviation

n=16 represent the sample size  

Part a

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $t_{\alpha/2}=1.96$

And the margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$  

And replacing we got:

$ME=1.96\frac{125}{\sqrt{16}}=61.25$  

c. 61.25 kg

Part b

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$    (a)

And on this case we have that ME =45 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (b)

$n=(\frac{1.960(125)}{45})^2 =29.64 \approx 30$

So the answer for this case would be n=30 rounded up to the nearest integer

b. 30

Part c

And for this case the conditions in order to use the confidence interval are satisfied since:

a. Yes. The distribution of sample means is normal because the data are normal.