Question

Assignment 2 Consider weights of all NBA players. Assume that we want to test the hypothesis that the average weight of the population of all players is 225 lbs, against the alternative hypothesis that the average weight is less that 225 lbs. Assume also that the population standard deviation of player weights is 26 lbs. Let the probability of making a type I error (a) be 0.05. What size random sample would be necessary if we want the probability of making a type Il error (B) to be 0.10, if the true population mean weight is 210 lbs. Draw a random sample of the size you computed above and compute the sample mean. Based on your sample result, would you reject of fail to reject the null hypothesis Given the hypothesis that you conducted, what would be the probability of making a typeIlerror if the actual mean weight were 210 lbs.__..

Answer 1

a)

Ho :   µ =   225  
Ha :   µ <   225  
          
True mean,   µ =    210  
hypothesis mean,   µo =    225  
          
Level of Significance ,    α =    0.05  
std dev =    σ =    26  

ß = 0.10   
δ=   µ - µo =    -15  
          
Z ( α ) =Z_0.05= 1.6449  
          
Z (ß) = Z_0.10 =   1.2816  
          
sample size needed =    n = ( σ [ Z(ß)+Z(α/2) ] / δ )² =    25.7296  
          
so, sample size =        26.000  
_______________________________

b)

let sample data be

208
272
295
194
168
166
177
292
183
227
180
289
166
175
279
258
208
285
215
223
158
220
193
181
182
258

Ho :   µ =   225
Ha :   µ <   225
      
Level of Significance ,    α =    0.05
population std dev ,    σ =    26
Sample Size ,   n =    26
Sample Mean,    x̅ =   217.3846154
      
'   '   '
      
Standard Error , SE =   σ/√n =   5.0990
      
Z-test statistic=   (x̅ - µ )/SE =    -1.4935
      
critical z value, z*   =   -1.6449
      
p-Value   =   0.0677
Conclusion:     p-value>α, Do not reject null hypothesis   
________________________

c)

true mean ,    µ =    210
      
hypothesis mean,   µo =    225
significance level,   α =    0.05
sample size,   n =   26
std dev,   σ =    26
      
δ=   µ - µo =    -15
Zα =       -1.6449   (left tail test)          
                      
We will fail to reject the null (commit a Type II error) if we get a Z statistic >               -1.6449      
this Z-critical value corresponds to X critical value( X critical), such that                      
                      
       (x̄ - µo)/σx ≥ Zα              
       x̄ ≥ Zα*σx + µo              
       x̄ ≥    216.613   (acceptance region)      
                      
now, type II error is ,ß =    P( x̄ ≥    216.613   given that µ =   210      
                      
   = P ( Z > (x̄-true mean)/σx )                   
   = P ( Z >    1.297   )          
   =   0.097              
                      
so, probability of type II error ,ß = 0.10