A nutrition expert claims that the average American is overweight. To test his c

Question

A nutrition expert claims that the average American is overweight. To test his claim, a random sample of 26 Americans was selected, and the difference between each person's actual weight and idea weight was calculated. For this data, we have x = 17.1 and s = 29.7. Is there sufficient evidence to conclude that the expert's claim is true? Carry out a hypothesis test at a 4% significance level. A. The value of the standardized test statistic (give to 3 decimal places): B. The p-value is (give to 4 decimal places): C. Your decision for the hypothesis test: A. Do Not Reject H_a. B. Do Not Reject H_0. C. Reject H_a. D. Reject H_0.

Explanation / Answer

Formulating the null and alternative hypotheses,              
              
Ho:   u   <=   0  
Ha:    u   >   0  
              
As we can see, this is a    right   tailed test.      
              
              
Getting the test statistic, as              
              
X = sample mean =    17.1
uo = hypothesized mean =    0          
n = sample size =    26
s = standard deviation =    29.7
              
Thus, z = (X - uo) * sqrt(n) / s =    2.935799114 [ANSWER]    = 2.936

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B)          
              
Also, the p value is, as this is right tailed,              
              
p =    0.001663    = 0.0017 [ANSWER]      
      
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As P < 0.04, we   REJECT THE NULL HYPOTHESIS.  

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