a survey is conducted among current MPH students to assess their knowledge of sa

Question

a survey is conducted among current MPH students to assess their knowledge of safe (alcohol) drinking limits. Two hundredstudents are randomly selected for the investigation, and 60% correctly identified safe drinking limits. a) construct a 95% confidence interval for the proportion of all MPH students who could correctly identify safe drinking limits. b) it has been reported that 70% of practicing clinicians can correctly identify safe drinking limits when asked in a survey format. one hundred of the MPH students involved in the survey are former or current practicing clinicians, and 72% of these individuals correctly identified safe drinking limits in the survey. is the proportionof current MPH students who are former or current practicing clinicians significantly different from the proportion reported in the literature? run the appropriate test at alpha = 0.05

Explanation / Answer

a)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.6          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.034641016          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.067895144          
lower bound = p^ - z(alpha/2) * sp =   0.532104856          
upper bound = p^ + z(alpha/2) * sp =    0.667895144          
              
Thus, the confidence interval is              
              
(   0.532104856   ,   0.667895144   ) [ANSWER]

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b)

Formulating the null and alternatuve hypotheses,          
          
Ho:   p   =   0.7
Ha:   p   =/=   0.7
As we see, the hypothesized po =   0.7      
Getting the point estimate of p, p^,          
          
p^ = x / n =    0.72      
          
Getting the standard error of p^, sp,          
          
sp = sqrt[po (1 - po)/n] =    0.045825757      
          
Getting the z statistic,          
          
z = (p^ - po)/sp =    0.43643578      
          
As this is a    2   tailed test, then, getting the p value,  
          
p =    0.662520584      

As P > 0.05, we   FAIL TO REJECT THE NULL HYPOTHESIS.      
Hence, the the proportion of current MPH students who are former or current practicing clinicians are not significantly different from the proportion reported in the literature (0.70). [CONCLUSION]

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