Question

The average height of a woman is 5'4. I asked 20 women 18 and older what their height was. One woman is 4'9, one is 5'1, one is 5'2, two are 5'3, four are 5'4, four are 5'5, four are 5'6, one is 5'8 and two are 5'9 For this sample, complete the following:

a. Calculate the mean, standard deviation, and variance.

b. Calculate the 95% Confidence Interval for the population mean.

c. Use your sample to test the hypothesis for the mean. Calculate the p-value for the test. Use α = 0.05 to find your conclusion.

d. Use your sample to test the hypothesis for the variance you proposed. Calculate the p-value for the test. Use α = 0.05 to find your conclusion.

Answer 1

a)

	X	(X - X̄)²
total sum	109.4	1.16
n	20	20

mean =    ΣX/n =    109.400   /   20   =   5.4700
                      
sample variance =    Σ(X - X̄)²/(n-1)=   1.1620   /   19   =   0.061
                      
sample std dev =   √ [ Σ(X - X̄)²/(n-1)] =   √   (1.162/19)   =       0.2473

................

b)

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 =   0.2473   / √   20   =   0.055298
margin of error , E=t*SE =   2.0930   *   0.05530   =   0.115741
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    5.47   -   0.115741   =   5.354259
Interval Upper Limit = x̅ + E =    5.47   -   0.115741   =   5.585741
95%   confidence interval is (   5.35   < µ <   5.59   )

...............

c)

Ho :   µ =   5.4                  
Ha :   µ ╪   5.4       (Two tail test)          
                          
Level of Significance ,    α =    0.05                  
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   0.2473                  
Sample Size ,   n =    20                  
Sample Mean,    x̅ = ΣX/n =    5.4700                  
                          
degree of freedom=   DF=n-1=   19                  
                          
Standard Error , SE = s/√n =   0.2473   / √    20   =   0.0553      
t-test statistic= (x̅ - µ )/SE = (   5.470   -   5.4   ) /    0.0553   =   1.27
                          
                          
p-Value   =   0.2209   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value>α, Do not reject null hypothesis                       
...............

THANKS

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