Question

(Heights of Presidents since 1900) Listed below are the heights (in inches) of the Presidents who started serving in 20th until today. Assume that these values are sample data from some larger population.

67, 70, 72, 71, 72, 70, 71, 74, 69, 70.5, 72,

75, 71.5   69.5, 73, 74, 74.5, 71.5, 73.5, 72.5

(a) Find Q₁ and Q₃

(b) Construct a boxplot for the data

(c) Find the range

(d) Find the standard deviation and variance

(e) Assume that the values are sample data, construct a 95% confidence interval to estimate of the population mean

(f) Assume that the values are sample data, construct a 95% confidence interval fro the population standard deviation

(g) The mean height of men is 69.0 inches. Use a 0.01 significance level to test the claim that this sample comes from a population with a mean greater than 69.0 inches. Do Presidents appear to be taller than a typical man in U. S.? (Show hypothesis test details, test statistic, decision making, and conclusion)

Answer 1

a)

quartile , Q1 = 0.25(n+1)th value=   5.25th   value of sorted data
=   70.125  
      
Quartile , Q3 = 0.75(n+1)th value=   15.75th   value of sorted data
=   73.375  

b)

c)

maximum =    75              
minimum=   67              
                  
range=max-min =    75   -   67   =   8

d)

sample variance =    Σ(X - X̄)²/(n-1)=   77.6375   /   19   =   4.086
                      
sample std dev =   √ [ Σ(X - X̄)²/(n-1)] =   √   (77.6375/19)   =       2.0214

e)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   19          
't value='   tα/2=   2.0930   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   2.0214   / √   20   =   0.452006
margin of error , E=t*SE =   2.0930   *   0.45201   =   0.946059
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    71.68   -   0.946059   =   70.728941
Interval Upper Limit = x̅ + E =    71.68   -   0.946059   =   72.621059
95%   confidence interval is (   70.73   < µ <   72.62   )

g)

Sample Size,   n=   20
Sample Standard Deviation,   s=   2.0214
Confidence Level,   CL=   0.95

confidence interval for std dev is       
lower bound=   √(lower bound variance)=   1.537
      
      
upper bound=   √(upper bound of variance=   2.952
      

g)

Ho :   µ =   69                  
Ha :   µ >   69       (Right tail test)          
                          
Level of Significance ,    α =    0.01                  
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   2.0214                  
Sample Size ,   n =    20                  
Sample Mean,    x̅ = ΣX/n =    71.6750                  
                          
degree of freedom=   DF=n-1=   19                  
                          
Standard Error , SE = s/√n =   2.0214   / √    20   =   0.4520      
t-test statistic= (x̅ - µ )/SE = (   71.675   -   69   ) /    0.4520   =   5.92
                          
  
                          
p-Value   =   0.0000   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value<α, Reject null hypothesis                       
Conclusion: There is enough evidence to that this sample comes from a population with a mean greater than 69.0 inches        

Thanks in advance!

revert back for doubt

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