0.4515 0.0948 0.4052 0.0485* e) 0.9515 We wish to give a 95% confidence for the

Question

0.4515 0.0948 0.4052 0.0485* e) 0.9515 We wish to give a 95% confidence for the lead content of water in a Michigan county. To this end a simple random sample water from 8 locations gives the following concentrations in parts per million. Y=1.4,13.2, 4.1, 4.3,4.5,1.0, 6.3, 7.4. Give theinargjn of error for_tte_95% confidence interval. 2.25 3.23^* 2.59 1.93 1.91 A poil is taken by obtaining a simple random sample of 500 adults. It is found that 270 members of the sample believe more money should be spent on medical research. Give a 99% confidence interval for the true proportion of adults who believe more money should be spent on medical research.

Explanation / Answer

Note that              
              
Margin of error = t(alpha/2) * s / sqrt(n)              
      
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025      
df = n - 1 = 8-1 =   7          
  
Hence, by table/technology,

t(alpha/2) = critical t for the confidence interval =    2.364624252          

Also, by technology,

s = sample standard deviation =    3.865506805  

Also,
      
n = sample size =    8          

Thus,              

Margin of error =    3.231644561 = 3.23 [ANSWER]

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