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You still know that adult female weight is normal with unknown mean and standard

ID: 3156938 • Letter: Y

Question

You still know that adult female weight is normal with unknown mean and standard deviation of 25 lbs. Your boss suspects that the mean adult female weight is 150 lbs You think that the mean adult female weight is not 150 lbs. State the appropriate hypotheses. After collecting a sample of 27 adult women and finding the sample mean to be 160 lbs. conduct the appropriate test at alpha .05 significance level. Do so also at alpha - .01 significance level Redo the previous test if you suspect that the mean adult female weight is greater than 150 lbs [ Redo the previous two hypotheses at each level of alpha if you suspect the mean adult female weight is different/less than 150 lbs and your calculated sample mean was 141 lbs ]

Explanation / Answer

A)

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   150  
Ha:    u   =/   150   [ANSWER]

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B)
              
As we can see, this is a    two   tailed test.      
Getting the test statistic, as              
              
X = sample mean =    160          
uo = hypothesized mean =    150          
n = sample size =    27          
s = standard deviation =    25          
              
Thus, z = (X - uo) * sqrt(n) / s =    2.078460969          
              
Also, the p value is, as this is two tailed,              
              
p =    0.0376669

Hence, at 0.05 level, as P < 0.05, we reject Ho.

At 0.05 level, there is significant evidence that the true mean adult female weight is not 150 lbs.

Meanwhile, at 0.01 level, as P > 0.01, we fail to reject Ho.  
              
At 0.01 level, there is no significant evidence that the true mean adult female weight is not 150 lbs.

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c)

Formulating the null and alternative hypotheses,              
              
Ho:   u   <=   150  
Ha:    u   >   150  
              
As we can see, this is a    right   tailed test.      
              
Getting the test statistic, as              
              
X = sample mean =    160          
uo = hypothesized mean =    150          
n = sample size =    27          
s = standard deviation =    25          
              
Thus, z = (X - uo) * sqrt(n) / s =    2.078460969          
              
Also, the p value is              
              
p =    0.018833461          

Hence, at 0.05 level, as P < 0.05, we reject Ho.

At 0.05 level, there is significant evidence that the true mean adult female weight is greater than 150 lbs.

Meanwhile, at 0.01 level, as P > 0.01, we fail to reject Ho.  
              
At 0.01 level, there is no significant evidence that the true mean adult female weight greater than 150 lbs.

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