1) The \"jumping seal\" lives in Coconut Island. Because this is arn endangered

Question

1) The "jumping seal" lives in Coconut Island. Because this is arn endangered species its population has been studied intensively Scientists have determined that the body size of adult jumping seals has a parametric mean of 1.83 m (measured from the tip of the snout to the tip of the tail) and a parametric standard deviation of 0.21 m. Recently, a single individual of jumping seal was found on a small island just south of Coconut Island. This single individual was very large, 2.13 m in length, and scientists think that this observation is very extreme a) Calculate the probability of finding an individual as extreme in size (at least 2.13 m) as the one found in the small island. b) The observation of the large individual has prompted scientists to visit the small island to study its population and determine whether it's really different in body size relative to the population in Coconut Island. To be able to answer this question they sampled 9 individuals from the small island and calculated a sample average of 2.11 m and a standard deviation of 0.15 m. Use a 95% confidence interval to determine whether there is a real indication that the small island population has different parametric mean compared to the one in Coconut Island. c) Calculate a 95% CI for the population variance of the small island population. Does the interval suggest that the two populations have a difference in variability for size?

Explanation / Answer

Solution:

Part a

We are given a mean = 1.83 and standard deviation = 0.21.

We have to find P(X2.13)

P(X2.13) = 1 – P(X<2.13)

Z = (X – mean)/SD

Z = (2.13 – 1.83)/0.21 = 1.428571

P(X<2.13) = P(Z< 1.428571) = 0.923436

P(X2.13) = 1 – P(X<2.13)

P(X2.13) = 1 – 0.923436

P(X2.13) = 0.076564

Required probability = 0.076564

Part b

We are given

Sample size = n = 9

Sample mean = Xbar = 2.11

Sample SD = 0.15

We have to find 95% confidence interval for population mean.

Confidence interval = Xbar -/+ t*SD/sqrt(n)

We are given

Confidence level = 95% (c = 0.95)

= 0.05, /2 = 0.025

Degrees of freedom = n – 1 = 9 – 1 = 8

Critical value = t = 2.3060

Confidence interval = 2.11 -/+ 2.3060*0.15/sqrt(9)

Confidence interval = 2.11 -/+ 2.3060*0.15/3

Confidence interval = 2.11 -/+ 0.1153

Lower limit = 2.11 - 0.1153 = 1.9947

Upper limit = 2.11 + 0.1153 = 2.2253

Confidence interval = (1.9947, 2.2253)

Above interval does not contain population mean 1.83, so there is a real indication that the small island population has different parametric mean compared to the one in Coconut Island.

Part c

Here, we have to find 95% CI for population variance of the small island population.

We are given

Sample size = n = 9

Sample mean = Xbar = 2.11

Sample SD = 0.15

Degrees of freedom = n – 1 = 9 – 1 = 8

Lower Chi square value = 2.1797

Upper Chi square value = 17.5345

Confidence interval = [(n – 1)*S^2/^2upper , (n – 1)*S^2/^2lower]

Lower limit = (9 – 1)*0.15^2/17.5345 = 8*0.15^2/17.5345 = 0.010265

Upper limit = (9 – 1)*0.15^2/2.1797 = 8*0.15^2/2.1797 =0.08258

Confidence interval = (0.010265, 0.08258)

Population variance = 0.21*0.21 = 0.0441

Population variance included in the above confidence interval, so this interval does not suggest that the two populations have a difference in variability for size.

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