2 A farmer is worried about the sweetness of his oranges fro
Solution
As the level of significance is not given, we assume it is 0.05.
Formulating the null and alternative hypotheses,              
               
 Ho:   u   >=   0.027  
 Ha:    u   <   0.027  
               
 As we can see, this is a    left   tailed test.      
               
 Thus, getting the critical z, as alpha =    0.05   ,      
 alpha =    0.05          
 zcrit =    -   1.644853627      
               
 Getting the test statistic, as              
               
 X = sample mean =    0.025          
 uo = hypothesized mean =    0.027          
 n = sample size =    30          
 s = standard deviation =    0.003          
               
 Thus, z = (X - uo) * sqrt(n) / s =    -3.651483717          
               
 Also, the p value is              
               
 p =    0.000130365          
               
 Comparing |z| > 1.645 (or, p < 0.05), we   REJECT THE NULL HYPOTHESIS.          
               
 Thus, there is significant evidence that the sugar content of the said grove is less than the sugar content of the other groves, at 0.027 lb per orange. [CONCLUSION]

