For each of the following studies the samples were given an
Solution
17.
FOR SAMPLE IN PART A, TWO TAILED Z TEST:
Formulating the null and alternative hypotheses,              
               
 Ho:   u   =   36  
 Ha:    u   =/   36  
               
 As we can see, this is a    two   tailed test.      
               
 Thus, getting the critical z, as alpha =    0.05   ,      
 alpha/2 =    0.025          
 zcrit =    +/-   1.959963985      
               
 Getting the test statistic, as              
               
 X = sample mean =    38          
 uo = hypothesized mean =    36          
 n = sample size =    16          
 s = standard deviation =    8          
               
 Thus, z = (X - uo) * sqrt(n) / s =    1          
               
 Also, the p value is              
               
 p =    0.317310508          
               
 As |z| < 1.95996, and P > 0.05, we   FAIL TO REJECT THE NULL HYPOTHESIS.          
Hence, there is no significant evidence that the true mean is 36, at 0.05 level. [CONCLUSION]
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FOR SAMPLE IN PART A, CONFIDENCE INTERVAL:
Note that              
 Margin of Error E = z(alpha/2) * s / sqrt(n)              
 Lower Bound = X - z(alpha/2) * s / sqrt(n)              
 Upper Bound = X + z(alpha/2) * s / sqrt(n)              
               
 where              
 alpha/2 = (1 - confidence level)/2 =    0.025          
 X = sample mean =    38          
 z(alpha/2) = critical z for the confidence interval =    1.959963985          
 s = sample standard deviation =    8          
 n = sample size =    16          
               
 Thus,              
 Margin of Error E =    3.919927969          
 Lower bound =    34.08007203          
 Upper bound =    41.91992797          
               
 Thus, the confidence interval is              
               
 (   34.08007203   ,   41.91992797   ) [ANSWER, CONFIDENCE INTERVAL]
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