Question

4) A 1992 article in the Journal of the American Medical Association ("A Critical Appraisal of 98.6 Degrees F, the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich") reported body temperature, gender, and heart rate for a number of subjects. The body temperatures for 25 female subjects follow: 97.8, 97.2, 97-4, 97.6, 97.8, 97.9, 98.0, 98.0, 98.0, 98.1, 98.2, 98.3, 98.3, 98.4, 98.4, 98.4, 98.5, 98.6, 98.6,98.7, 98.8, 98.8, 98.9, 98.9, and 99.0. Test the hypothesis H: u = 98.6 versus H: # 98.6, using a = 0.05. Find the P-value. tol = 3.48 > 2.064 reject H P-value = 0.002 Check the assumption that female body temperature is normally distributed. Yes Compute the power of the test if the true mean female body temperature is as low as 98.0. power 21 What sample size would be required to detect a true mean female body temperature as low as 98.2 if you wanted the power of the test to be at least 0.9? 71 = 20

*Show ALL work, answer is given already*

Answer 1

Ho :   µ =   98.6  
Ha :   µ ╪   98.6   (Two tail test)
          
Level of Significance ,    α =    0.050  
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   0.4821  
Sample Size ,   n =    25  
Sample Mean,    x̅ = ΣX/n =    98.2640  
          
Standard Error , SE = s/√n =   0.4821/√25=   0.0964  
t-test statistic= (x̅ - µ )/SE =    (98.264-98.6)/0.0964=   -3.48
          
degree of freedom=   DF=n-1=   24  
critical t value, t* =    ±   2.064 [Excel formula =t.inv(α/2,df) ]
          
p-Value   =   0.002   [Excel formula =t.dist.2t(t-stat,df) ]
Decision:   p-value≤α, Reject null hypothesis       

b)

Probability Plot of Raw Data Normal - 95% CI 99 Mean StDev N AD P-Value 95 98.26 0.4821 25 0.238 0.759 90 80 70 60 Percent 50 40 30 20 10 5 1 97.0 97.5 98.0 98.5 99.0 99.5 100.0 Raw Data

probability plot from MINITAB, we can observe that data is normally distributed

c)

true mean ,    µ =    98
      
hypothesis mean,   µo =    98.6
significance level,   α =    0.05
sample size,   n =   25
std dev,   σ =    0.4821
      
δ=   µ - µo =    -0.6
      
std error of mean=σx = σ/√n =    0.4821/√25=   0.0964

(two tailed test) Zα/2   = ±   1.960      
We will fail to reject the null if we get a Z statistic between              
-1.960   and   1.960      
these Z-critical value corresponds to some X critical values, such that              
-1.960   ≤(x̄ - µo)/σx≤   1.960      
98.411   ≤ x̄ ≤   98.789      
now, type II error is ,              
ß = P (   98.411   ≤ x̄ ≤   98.789   )
Z =    (x̄-true mean)/σx          
Z1=(98.411-98)/0.0964=       4.2631      
Z2=(98.789-98)/0.0964=       8.1830      
              
P(Z<8.184)-P(Z<4.2641)=              
=   1   -   1 = 0
Power = 1 - ß=   1          

d)

true mean= µ =   98.2  
hypothesized mean=µo =    98.6  
α=   0.05  
std dev,σ=   0.4821  
power= 1- ß =   0.9  
ß =    0.1  
δ=µ - µo =    0.4  
Zα/2=   1.9600  
Z (ß ) =    1.2816  
n = ( ( Z(ß)+Z(α) )*σ / δ )² =    ((1.282+1.96)*0.4821/0.4)^2=   15.26
so, sample size=   16