The number of pups with wolf dens of the southwestern Unitedstates is recorded b

Question

The number of pups with wolf dens of the southwestern Unitedstates is recorded below for 16 wolf dens. (The Wolf in thesouthwest: The making of an endangered species, edited byD.E.Brown, University of Arizona Press). 5   8   7   5  3   4   3   9 5   8   5   6  5   6   4   7 a. Use a calculator with mean and standard deviation keys toverify that the sample mean is _ ˜ 5.63 pups with samplestandard deviation s ˜ 1.78pups.                                                           x b. Compute an 85% confidence interval for the population meannumber of wolf pups per den in the southwestern UnitedStates. The number of pups with wolf dens of the southwestern Unitedstates is recorded below for 16 wolf dens. (The Wolf in thesouthwest: The making of an endangered species, edited byD.E.Brown, University of Arizona Press). 5   8   7   5  3   4   3   9 5   8   5   6  5   6   4   7 a. Use a calculator with mean and standard deviation keys toverify that the sample mean is _ ˜ 5.63 pups with samplestandard deviation s ˜ 1.78pups.                                                           x b. Compute an 85% confidence interval for the population meannumber of wolf pups per den in the southwestern UnitedStates.

Explanation / Answer

The numberof pups with wolf dens of the southwestern United states is
recorded below for 16 wolf dens.
5   8   7   5  3   4   3  9 5   8   5   6  5   6   4  7

a. Use acalculator with mean and standard deviation keys to verify that
the sample mean is X_bar ˜ 5.63 pups with sample standarddeviation s ˜ 1.78 pups.
   {SampleMean} = (X_bar) = 5.625
     {Sample Standard Deviation} = s = 1.784

b. Computean 85% confidence interval for the population mean number of
wolf pups per den in the southwestern UnitedStates.
   {Sample Mean} = (X_bar) = 5.625
     {Sample Standard Deviation} = s = 1.784
     {Sample Size} = N = 16
     {Degrees Of Freedom} = df = N - 1 = (16) - 1 = 15
     {"t" Value For 85% Confidence Interval,df=15} = (t_85) = 1.517
     {85%Confidence Interval For Population Mean""} =
          = { (X_bar) -(t_85)*s/Sqrt[N] < < (X_bar) + (t_85)*s/Sqrt[N] }
          = { (5.625)- (1.517)*(1.784)/Sqrt[16] < < (5.625)+ (1.517)*(1.784)/Sqrt[16] }
          = { (4.948) < < (6.302)}

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