Question

Diastolic blood pressure for diabetic women has a normal distribution with unknown mean and a standard deviation equal to 10 mmHg. Researchers want to know if the mean DBP of diabetic women is equal to the mean DBP among the general public, which is known to be 76 mmHg. A sample of 10 diabetic women is selected and their mean DBP is calculated as 85mmHg.

a. Conduct the appropriate hypothesis test at the 0.01 significance level.

b. What would a Type-1 error in example setting be?

c. How much power do you have to detect a difference of 11 mmHg between men and women?

Answer 1

a)

Ho :   µ =   76  
Ha :   µ ╪   76  
          
Level of Significance ,    α =    0.01  
population std dev ,    σ =    10  
Sample Size ,   n =    10  
Sample Mean,    x̅ =   85  
          

          
Standard Error , SE =   σ/√n =   3.1623  
          
Z-test statistic=   (x̅ - µ )/SE =    2.8460  
          
critical z value, z*   =   2.5758   [Excel formula =NORMSINV(α/no. of tails) ]
          
p-Value   =   0.0044  
Conclusion:     p-value<α, Reject null hypothesis

so, there is enough evidence to conlcude that mean DBP of diabetic women is different  to the mean DBP among the general public which is 76 mmHg

b)

Type I error is rejecting null hypothesis when it is true

so, here type I error is to conclude that mean DBP of diabetic women is different  to the mean DBP among the general public but in actual mean DBP of diabetic women is equal to the mean DBP among the general public

c)

hypothesis mean,   µo =    76  
significance level,   α =    0.01  
sample size,   n =   10  
std dev,   σ =    10  
          
δ=   µ - µo =    11  
          
std error of mean,   σx = σ/√n =    3.1623  
          
Zα/2   = ±   2.576   (two tailed test)

ß =   P(Z < Zα/2 - δ√n/σ) - P(Z < -Zα/2-δ√n/σ) =P(Z<( 2.576 - 11/3.1623)) - P(Z < -2.576 - 11/3.1623) =  
   = P ( Z <    -0.9027   ) - P ( Z <   -6.0543   )
                  
   =   0.1833   -    0.0000  
   =   0.1833     

    power =    1 - ß =   0.8167  

so, power is 0.8167