IN example 111 page 266 we saw that the distribution of heig

IN example 11.1 (page 266) we saw that the distribution of heights of young women aged 18 to 24 is approx. Normal with mean =64.5 in and standard deviation =2.5 in. A random sample of 93 female undergraduates at UC Irvine produced a sample mean height of 62.7 in. Is there evidence that the mean height of all female undergraduates is different from the mean height of all young women in the US? Follow the 4 step process.

Solution

We assumed 0.05 = alpha as it is not given here.

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Formulating the null and alternative hypotheses,              
              
Ho:   u   =   64.5  
Ha:    u   =/   64.5  
              
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 =    62.7          
uo = hypothesized mean =    64.5          
n = sample size =    93          
s = standard deviation =    2.5          
              
Thus, z = (X - uo) * sqrt(n) / s =    -6.943428548          
              
Also, the p value is              
              
p =    3.82696*10^-12          
              
As |z| > 1.95996, or P < 0.05, we   REJECT THE NULL HYPOTHESIS.          
              
Thus, there is significant evidence that the mean height of all female undergraduates at UC Irvine is different from the mean height of all young women in the US. [CONCLUSION]