1. Gravetter/Wallnau Easentials-Chapter &-End-ef-chapter question 6 rardom sampl

Question

1. Gravetter/Wallnau Easentials-Chapter &-End-ef-chapter question 6 rardom sample is seleted mom a romal ppdatin with . me" of ' and a sanar dniatin of 0-8 Ater a trestment is administered to the indiiduels in tte ample, the sampe mean is fhund be be M- f the sample consists ofn 16 scores, is the smple mean sumclent t to concude that the treatment has asigriicant condude that the treetment hws a grificant effect. · et scoreS, a the soripla ioas sumcient so conclude Bhat the treatment has 3 sgrincant tect? use a tvo taded beal wih- .o5

Explanation / Answer

3.

If n = 16:

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   30  
Ha:    u   =/   30  
              
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 =    33          
uo = hypothesized mean =    30          
n = sample size =    16          
s = standard deviation =    8          
              
Thus, z = (X - uo) * sqrt(n) / s =    1.5   [ANSWER]      
              
Also, the p value is              
              
p =    0.133614403          
              
As |z| < 1.96, and P > 0.05, we   FAIL TO REJECT THE NULL HYPOTHESIS.          

You CANNOT conclude that the treatment has a significant effect. [ANSWER]

****************************************************

If n = 64:

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   30  
Ha:    u   =/   30  
              
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 =    33          
uo = hypothesized mean =    30          
n = sample size =    64          
s = standard deviation =    8          
              
Thus, z = (X - uo) * sqrt(n) / s =    3   [ANSWER, Z]      
              
Also, the p value is              
              
p =    0.002699796          
              
As |z| > 1.96, and P < 0.05, we   FAIL TO REJECT THE NULL HYPOTHESIS.          

You CAN conclude that the treatment has a significant effect. [ANSWER]

****************************************************

Hence,

A larger sample INCREASES the likelihood of rejecting the null hypothesis. [ANSWER]

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