A sample of 41 lots in a suburb has a mean of 192 nesting bi

A sample of 41 lots in a suburb has a mean of 19.2 nesting birds with a standard deviation of 3.5 birds. e. With .05 significance, test the claim that the standard deviation in number of nesting birds is 5 birds. Use the critical value (traditional) method.

f. If the population standard deviation is known to be 3.8 birds, test the claim that the mean number of nesting birds in this suburb is greater than 18. Use the P-value method with formula & chart.

Solution

Formulating the null and alternative hypotheses,              
              
Ho:   u   <=   18  
Ha:    u   >   18  
              
As we can see, this is a    right   tailed test.      
              
              
Getting the test statistic, as              
              
X = sample mean =    19.2          
uo = hypothesized mean =    18          
n = sample size =    41          
s = standard deviation =    3.8          
              
Thus, z = (X - uo) * sqrt(n) / s =    2.022039233          
              
Thus, the right tailed p value is              
              
p =    0.021586149          
              
As P < 0.05, we   REJECT THE NULL HYPOTHESIS.      

Thus, there is significant evidence that the mean number of nesting birds in this suburb is greater than 18. [CONCLUSION]