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How much does a wild mountain lion weigh? Bill Montana went into the mountains of Idaho and captured 12 random mountain lions. Their weights, in pounds, were the following:

93.0, 109.4, 88.2, 114.5, 83.0, 103.2, 110.7, 128.9, 62.8, 97.9, 94.7, 92.5

Mountain lion weights in Idaho are known to be normally distributed with an unknown mean of [tex]\mu[/tex] pounds and an unknown standard deviation [tex]\sigma[/tex] pounds.

a) Calculate the sample mean for this data.

b) Calculate the sample standard deviation for this data.

c) Calculate the maximum likelihood estimate for [tex]\sigma^2[/tex] using this data.

d) Calculate an unbiased estimate for [tex]\sigma^2[/tex] using this data.

e) Calculate the sample median for this data.

f) Suppose [tex]W[/tex] has a t distribution with 11 degrees of freedom. If [tex]P(W > t) = 0.08[/tex], then what is [tex]t[/tex]?

g) Suppose [tex]W[/tex] has a t distribution with 11 degrees of freedom. If [tex]P(W < t) = 0.08[/tex], then what is [tex]t[/tex]?

h) Calculate the 90th percentile of a standard normal distribution.

i) Compute an 80% confidence interval for [tex]\mu[/tex] using your answers above.

j) Compute an 80% prediction interval for [tex]\mu[/tex] using your answers above.

k) Copy your R script or other comments for the above into the text box here.

Answer :

(a) The sample mean is 98.23.

(b) The standard deviation is 16.97.

(c) The maximum likelihood estimate is 264.24.

(d) An unbiased estimate is 288.26.

(e) The sample median is 96.3

Bill Montana went into the mountains of Idaho and captured 12 random mountain lions. Their weights, in pounds, were the following:

93.0, 109.4, 88.2, 114.5, 83.0, 103.2, 110.7, 128.9, 62.8, 97.9, 94.7, 92.5

a) We have to find the sample mean.

Sample mean = Sum of all weights/ Total number

Sample mean = 1178.8/12

Sample mean = 98.23

b) We have to find the standard deviation.

The formula is

Standard Deviation(SD) = √{∑(Mean - Weight)^2/ Total number}

After using the formula

Sample standard deviation (SD) = 16.97

c) We have to calculate the maximum likelihood estimate.

The MLE of sigma^2 is given by:

sigma^2_(MLE) = 1/nsum_(i = 1)(x_i-barx)^2

by calculations we get ,

sigma^2_(MLE) = 264.24

d) We have to calculate an unbiased estimate.

The unbiased estimator of sigma^2 " is"

sigma^2_(UE) = 1/(n-1)sum_(i = 1)(x_i-barx)^2

By calculations, we get:

sigma^2_(UE) = 288.26

e) Now we have to calculate the sample median.

The sample median = 96.3.

h) Now we have to calculate the 90th percentile of a standard normal distribution.

let, the 90th percentile of standard normal distribution be "k"

then ,

P(X < k) = 0.90

=> P(X < k) = P(X < 1.281)

=> k = 1.281

Therefore, the 90th percentile of standard normal distribution is 1.281.

(i) Now we have to find an 80% confidence interval.

The upper bound of confidence interval of is by calculations , we get 101.91.

The lower bound of confidence interval of is by calculations, we get 94.54.

Therefore, the confidence interval is : (94.54 , 101.91)

To learn more about confidence interval link is here

brainly.com/question/24131141

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