Question

In: Statistics and Probability

Adults aged 18 years old and older were randomly selected for a survey on obesity. Adults...

Adults aged 18 years old and older were randomly selected for a survey on obesity. Adults are considered obese if their body mass index (BMI) is at least 30. The researchers wanted to determine if the proportion of women in the south who are obese is less than the proportion of southern men who are obese. The results are shown in the table below.

# who are obese sample size
Men 42,769 155,525
Women 67,169 248,775

At the1% level of significance, is there evidence to support the claim? Explain.

work must include

1. Clear statement of hypotheses, with the correct parameter(s)
2. An indication of the test used
3. The test statistic and p-value
4. An indication of the statistical decision (i.e. whether or not to reject Ho)
     along with an explanation.
5. An interpretation of the statistical decision in the context of the problem.  

Solutions

Expert Solution

1)

Ho: proportion of women in the south who are obese is equal to the proportion of southern men who are obese   
Ha: proportion of women in the south who are obese is less than the proportion of southern men who are obese  

Ho:   p1 - p2 =   0          
Ha:   p1 - p2 >   0  

2)

Two proportion z test

  
3)


sample #1   ----->   Men   
first sample size,     n1=   155525          
number of successes, sample 1 =     x1=   42769          
proportion success of sample 1 , p̂1=   x1/n1=   0.2750          
                  
sample #2   ----->   Women   
second sample size,     n2 =    248775          
number of successes, sample 2 =     x2 =    67169          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.270          
                  
difference in sample proportions, p̂1 - p̂2 =     0.2750   -   0.2700   =   0.0050
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.2719          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0014          
Z-statistic = (p̂1 - p̂2)/SE = (   0.005   /   0.0014   ) =   3.4753
                  

p-value =        0.0003   [excel function =NORMSDIST(-z)]      

4)

decision :    p-value<α,Reject null hypothesis               
                  
5)

Conclusion:   There is enough evidence to support that the proportion of women in the south who are obese is less than the proportion of southern men who are obese.

Please revert back in case of any doubt.

Please upvote. Thanks in advance.


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