Mathematics, 10.03.2020 03:44 mcalepcrager

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.

Answers: 1

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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.

Answer from: Quest

$x=\frac{3}{4}$

step-by-step explanation:

we are given

$-6+2x=6x-9$

since, we have to isolate x on anyone side

so, we can move all x terms on one side

subtract both sides by 2x

$-6+2x-2x=6x-9-2x$

$-6=4x-9$

add both sides by 9

$-6+9=4x-9+9$

$3=4x$

divide both sides by 4

we get

$x=\frac{3}{4}$

Answer from: Quest

c. x=2.4cm

step-by-step explanation:

similar triangles are triangles which have the same shape but not the same size. this means their angle measures are equal but their side lengths are not instead they are proportional. to solve, we will set up a proportion and solve.

a proportion is two equal ratios set equal to each other. we form the ratios by finding corresponding sides (sides which match to each other on both triangles by their position). out ratios will be one side of the small triangle over the corresponding side on the big triangle as shown below:

$\frac{little}{big} =\frac{side mn}{side ab}=\frac{side on}{side cb}$ .

$\frac{0.96}{1.2}=\frac{x}{3}$

to solve the proportion, we'll cross multiply by multiplying numerator with denominator across the equal sign.

$0.96(3)=1.2(x)\\2.88=1.2x\\\frac{2.88}{1.2}=\frac{1.2x}{1.2} \\2.4=x$

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