Deer mice Peromyscus maniculatus are small rodents native to

Deer mice (Peromyscus maniculatus) are small rodents native to North America. Their adult body lengths (excluding tail) are known to vary approximately Normally, with mean mu = 86 millimeters (mm) and standard deviation sigma = 8 mm. 9 Deer mice are found in diverse habitats and exhibit different adaptations to their environment. A random sample of 14 deer mice in a rich forest habitat gives an average body length of = 91.1 mm. Assume that the standard deviation sigma of all deer mice in this area is also 8 mm. 9 YOUR FORMULAS AND ALL WORK ON THIS PAGE a) What is the standard deviation of the mean length? Standard Deviation b) What critical value do you need to use in order to compute a 95% confidence interval for the 16 mean mu? Critical Value (Z-score) c) Give a 95% confidence interval for the mean body length of all deer mice in the forest habitat. Lower Bound of Confidence Interval Upper Bound of Confidence Interval d) Give a 90% confidence interval for the mean body length of all deer mice in the forest habitat. Lower Bound of Confidence Interval Upper Bound of Confidence Interval e) This confidence interval is shorter than your interval in the previous exercise. even though the 35 Intervals come from the same sample. Why does the second interval have a smaller margin of 36 error? Explain

Solution

a)

s(X) = s/sqrt(n) = 8/sqrt(9) = 2.666666667 [ANSWER]

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b)

alpha/2 = (1 - confidence level)/2 =    0.025          
z(alpha/2) = critical z for the confidence interval =    1.959963985   [ANSWER]

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c)      

Note that              

Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    91.1          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    8          
n = sample size =    9          
              
Thus,              

Lower bound =    85.87342937          
Upper bound =    96.32657063   [ANSWERS]

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d)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.05          
X = sample mean =    91.1          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    8          
n = sample size =    9          
              
Thus,              

Lower bound =    86.71372366          
Upper bound =    95.48627634   [ANSWERS]      
  
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Hi! Please attach the previous exercise and resubmit this question to get the answer to part E. Thanks!