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 LaTeX: \sigma = 125????=125 kg, and the researchers feel comfortable using this standard deviation for their confidence interval.
Use this information to answer Questions.
. 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.
15.31 kg
51.40 kg
61.25 kg
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?
25
30
35
40
Are the requirements for the use of a confidence interval met? Explain.
Yes. The distribution of sample means is normal because the data are normal.
Yes. The distribution of sample means is normal because the sample size is large.
No. The distribution of sample means is not normal because the sample size is small.
No. The fact that the data are normal does not imply that the distribution of sample means is normal.

Answer 1

Answer: a) margin of error = 61.25, b) sample size when margin of error is 45 = 30

Step-by-step explanation:

The formulae to get the margin of error of a confidence interval is given as

Margin of error = critical value * (σ/√n)

Where σ = population standard deviation = 125

n = sample size = 16

Critical value =Zα/2 = 1.96 ( this is so because we are performing a 95% confidence level test then level of significance (α) will be 5% and since our test is of two values, it will be 2 tailed).

Margin of error = 1.96 * (125/√16)

Margin of error = 1.96 * 125/4

Margin of error = 1.96 * 31.25

Margin of error = 61.25

Question b)

Margin of error = 45

Critical value =Zα/2 = 1.96

Population standard deviation = σ = 125

Sample size =n =??

By recalling the formulae

Margin of error = critical value * (σ/√n)

45 = 1.96 * (125/√n)

45 = (1.96 * 125)/√n

45 = 245/√n

45 * √n = 245

√n = 245/ 45

√n = 5.444

n = (5.444)²

n= 29.64 which is approximately 30.